## 1. Abridged Problem Statement

Bencoding represents four kinds of objects:

- **String**: `k:s`, where `k` is the length of `s`, written without leading zeros except `0`.
- **Integer**: `i<n>e`, where `n` is a non-negative integer without leading zeros except `0`.
- **List**: `l<item1><item2>...e`
- **Dictionary**: `d<key1><value1><key2><value2>...e`

Lists and dictionaries may be empty. Dictionary keys and values can be any bencoded object.

Given:

- an integer `m`, the maximum allowed length of a valid bencoded object,
- a string `c`,

determine whether `c` itself is a valid single bencoded object of length at most `m`.

If yes, print:

```text
ok
```

Otherwise, find the maximum position `j` such that `c[0:j]` can still be extended into some valid bencoded object of total length at most `m`, and print:

```text
Error at position j!
```

Positions are zero-based, and `j` is also the prefix length.

---

## 2. Detailed Editorial

We need to parse a possibly incomplete bencoded string and determine the longest prefix that can be extended to a valid complete bencoded object of total length at most `m`.

A naive recursive parser is not enough because we also need to know, after every prefix, whether it is still possible to finish the object within the length limit `m`.

The provided solution performs one left-to-right scan using a state machine.

---

### Parsing states

While scanning, we maintain two kinds of state.

#### 1. Stack of open containers

The stack stores currently open lists and dictionaries.

For dictionaries, we need to know whether we are currently expecting a key or a value.

There are three possible frame types:

```cpp
LIST
DICT_K  // dictionary currently expecting a key, or it may close
DICT_V  // dictionary currently expecting a value, cannot close yet
```

For example:

- In a list, at a boundary, either another item may start or the list may end with `e`.
- In a dictionary expecting a key, either a key may start or the dictionary may end with `e`.
- In a dictionary expecting a value, an item must start; `e` is illegal.

---

#### 2. Current partial object state

Besides containers, we may be in the middle of reading an integer or string.

There are four possibilities:

```cpp
NONE      // currently at a clean object boundary
INT       // after reading 'i', reading integer digits until 'e'
STR_LEN   // reading string length digits until ':'
STR_BODY  // reading raw string body characters
```

For integers and string lengths, we also track leading-zero validity.

---

### Completing a parsed item

Whenever an item finishes:

- If there is no open container, then the root object has completed.
- If the parent is a dictionary expecting a key, it now expects a value.
- If the parent is a dictionary expecting a value, it now expects another key or closing `e`.
- If the parent is a list, nothing changes; another list item may follow or the list may close.

---

### Detecting invalid syntax

During parsing, a prefix becomes invalid immediately if, for example:

- an integer has no digits before `e`,
- an integer or string length has illegal leading zeros,
- an unexpected character appears,
- a dictionary expecting a value receives `e`,
- characters appear after the root object is already complete.

When this happens, scanning stops.

---

### Length feasibility

Even if the prefix is syntactically possible, it may be impossible to finish within length `m`.

So after every successfully consumed character, we compute the minimum number of extra characters needed to finish a valid object.

If:

```text
current_prefix_length + minimum_remaining_length > m
```

then this prefix cannot be extended into a valid object within the limit, so scanning stops.

---

### Minimum remaining length

The key part of the solution is computing the shortest possible suffix that completes the current prefix.

The smallest complete bencoded object has length `2`, for example:

```text
0:
le
de
```

So before anything starts, the minimum remaining length is `2`.

---

### Container closing costs

For each open frame, we maintain a value called `g_cost`.

This is the minimum number of characters needed to finish that frame assuming the currently active item inside it has just completed.

The values are:

```cpp
LIST   -> 1
DICT_K -> 3
DICT_V -> 1
```

Why?

#### LIST

After an item inside a list finishes, we can close the list with:

```text
e
```

Cost: `1`.

#### DICT_K

If a dictionary is expecting a key, and we are currently parsing that key, then after the key finishes, the dictionary will require a value before closing.

The shortest value is an empty string:

```text
0:
```

Then close dictionary:

```text
e
```

So cost:

```text
0:e
```

Length `3`.

#### DICT_V

If a dictionary is expecting a value, and that value finishes, then the dictionary can close immediately with:

```text
e
```

Cost `1`.

---

### Clean-boundary costs

When we are at a clean boundary inside a frame, the minimum cost is slightly different.

We define `f_cost`:

```cpp
LIST   -> 1
DICT_K -> 1
DICT_V -> 3
```

Why?

- Inside a list, we can close with `e`.
- Inside a dictionary expecting a key, we can close with `e`.
- Inside a dictionary expecting a value, we must provide a shortest value `0:` and then close with `e`, total length `3`.

---

### Maintaining `g_sum`

Instead of recomputing costs over the entire stack each time, the solution maintains:

```cpp
g_sum = sum of g_cost(frame) over all open frames
```

This allows `min_remaining()` to run in `O(1)` time.

---

### Computing minimum remaining length

There are two main cases.

---

#### Case 1: We are at a clean boundary

If no root object has started:

```cpp
return 2;
```

If root is completed:

```cpp
return 0;
```

If inside a container, then:

```cpp
f_cost(top_frame) + g_sum - g_cost(top_frame)
```

That means:

- finish or close the current innermost frame using `f_cost`,
- then finish all outer frames using their `g_cost`.

---

#### Case 2: We are in the middle of an item

First compute how many characters are needed to finish that item:

- Integer:
  - if no digits yet: need one digit plus `e`, cost `2`
  - otherwise: need just `e`, cost `1`
- String length:
  - need `:` plus exactly `length` body characters
- String body:
  - need remaining body characters

Then add `g_sum`, because after this item completes, every open container still needs to be completed.

---

### Final answer

We track:

```cpp
j_max
```

the largest prefix length seen so far that is both:

1. syntactically valid so far,
2. completable within length `m`.

At the end:

- if the whole string was consumed,
- and the root object is completed,

then print `ok`.

Otherwise print:

```text
Error at position j_max!
```

The algorithm is linear:

```text
O(|c|)
```

Memory usage is proportional to nesting depth:

```text
O(|c|)
```

---

## 3. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int m;
string c;

void read() { cin >> m >> c; }

void solve() {
    // Single linear scan with a state machine. The state has two parts:
    //   frames   - stack of open containers, each LIST, DICT_K (boundary inside
    //              a dict, expecting next key item or 'e') or DICT_V (just saw
    //              a key, must read a value item next; 'e' not allowed)
    //   progress - what we're in the middle of parsing: NONE (at a clean
    //              boundary), INT (digits then 'e'), STR_LEN (digits then ':',
    //              accumulated length in str_val) or STR_BODY (str_val chars of
    //              raw body left to consume)
    //
    // To answer "can the prefix be completed within m chars?" we maintain
    // g_sum, the sum of per-frame g_cost values for all open frames. g_cost is
    // the cost to close a frame assuming whatever sits inside it will complete
    // on its own (just needs the trailing chars): 1 for LIST ('e'), 3 for
    // DICT_K ("0:" filler value + 'e'), 1 for DICT_V ('e' after the pending
    // value finishes). f_cost is the cost to close from a clean boundary
    // *inside* that frame, with no item lined up: 1 for LIST/DICT_K, 3 for
    // DICT_V (must add "0:" then 'e'). Then min_remaining is finish_cost +
    // g_sum when an item is in progress, or f_cost(top) + (g_sum -
    // g_cost(top)) at a clean boundary. Special case: empty stack and no item
    // started yet needs 2 chars for the smallest object.
    //
    // A prefix length j is feasible iff parsing c[0..j-1] never violated a rule
    // and j + min_remaining <= m; we track the largest such j, and print "ok"
    // only when the whole string parsed and the root object closed.

    enum Frame { LIST, DICT_K, DICT_V };
    enum Progress { NONE, INT, STR_LEN, STR_BODY };

    vector<Frame> frames;
    Progress progress = NONE;
    int int_digits = 0;
    bool int_zero = false;
    int64_t str_val = 0;
    int str_len_digits = 0;
    bool str_len_zero = false;
    bool root_started = false;
    bool root_completed = false;
    int64_t g_sum = 0;

    auto g_cost = [](Frame f) -> int { return (f == DICT_K) ? 3 : 1; };

    auto f_cost = [](Frame f) -> int { return (f == DICT_V) ? 3 : 1; };

    auto min_remaining = [&]() -> int64_t {
        if(root_completed) {
            return 0;
        }

        if(progress == NONE) {
            if(!root_started) {
                return 2;
            }

            if(frames.empty()) {
                return 0;
            }

            Frame top = frames.back();
            return f_cost(top) + (g_sum - g_cost(top));
        }

        int64_t finish_cost = 0;
        if(progress == INT) {
            finish_cost = (int_digits == 0 ? 1 : 0) + 1;
        } else if(progress == STR_LEN) {
            finish_cost = 1 + str_val;
        } else {
            finish_cost = str_val;
        }

        return finish_cost + g_sum;
    };

    auto item_completed = [&]() {
        if(frames.empty()) {
            root_completed = true;
        } else {
            Frame& top = frames.back();
            if(top == DICT_K) {
                g_sum -= g_cost(DICT_K);
                top = DICT_V;
                g_sum += g_cost(DICT_V);
            } else if(top == DICT_V) {
                g_sum -= g_cost(DICT_V);
                top = DICT_K;
                g_sum += g_cost(DICT_K);
            }
        }
    };

    auto start_item = [&](char ch) -> bool {
        if(ch == 'i') {
            progress = INT;
            int_digits = 0;
            int_zero = false;
            return true;
        }

        if(ch == 'l') {
            frames.push_back(LIST);
            g_sum += g_cost(LIST);
            return true;
        }

        if(ch == 'd') {
            frames.push_back(DICT_K);
            g_sum += g_cost(DICT_K);
            return true;
        }

        if(ch >= '0' && ch <= '9') {
            progress = STR_LEN;
            str_val = ch - '0';
            str_len_digits = 1;
            str_len_zero = (ch == '0');
            return true;
        }

        return false;
    };

    auto process = [&](char ch) -> bool {
        if(root_completed) {
            return false;
        }

        if(progress == INT) {
            if(ch >= '0' && ch <= '9') {
                if(int_zero && int_digits >= 1) {
                    return false;
                }

                if(ch == '0' && int_digits == 0) {
                    int_zero = true;
                }

                int_digits++;
                return true;
            }

            if(ch == 'e') {
                if(int_digits == 0) {
                    return false;
                }

                progress = NONE;
                item_completed();
                return true;
            }

            return false;
        }

        if(progress == STR_LEN) {
            if(ch >= '0' && ch <= '9') {
                if(str_len_zero && str_len_digits >= 1) {
                    return false;
                }

                str_val = str_val * 10 + (ch - '0');
                str_len_digits++;
                return true;
            }

            if(ch == ':') {
                progress = STR_BODY;
                if(str_val == 0) {
                    progress = NONE;
                    item_completed();
                }

                return true;
            }

            return false;
        }

        if(progress == STR_BODY) {
            str_val--;
            if(str_val == 0) {
                progress = NONE;
                item_completed();
            }

            return true;
        }

        if(frames.empty()) {
            if(root_started) {
                return false;
            }

            root_started = true;
            return start_item(ch);
        }

        Frame top = frames.back();
        if(top == LIST || top == DICT_K) {
            if(ch == 'e') {
                g_sum -= g_cost(top);
                frames.pop_back();
                item_completed();
                return true;
            }

            return start_item(ch);
        }

        if(ch == 'e') {
            return false;
        }

        return start_item(ch);
    };

    int n = (int)c.size();
    int j_max = 0;

    for(int i = 0; i < n; i++) {
        if(!process(c[i])) {
            break;
        }

        int64_t new_j = i + 1;
        if(new_j + min_remaining() > (int64_t)m) {
            break;
        }

        j_max = (int)new_j;
    }

    if(j_max == n && root_completed) {
        cout << "ok\n";
    } else {
        cout << "Error at position " << j_max << "!\n";
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4. Python Solution

```python
import sys


def solve() -> None:
    # Read all input lines.
    data = sys.stdin.read().splitlines()

    # First line: maximum allowed valid-object length.
    m = int(data[0])

    # Second line: character sequence to process.
    c = data[1]

    # Frame types for the stack of open containers.
    LIST = 0      # Inside a list.
    DICT_K = 1    # Inside a dictionary, expecting a key or closing 'e'.
    DICT_V = 2    # Inside a dictionary, expecting a value.

    # Parser progress states.
    NONE = 0      # At a clean boundary.
    INT = 1       # Reading integer digits after 'i'.
    STR_LEN = 2   # Reading string length digits.
    STR_BODY = 3  # Reading string body characters.

    # Stack of open containers.
    frames = []

    # Current parser progress.
    progress = NONE

    # Integer parsing state.
    int_digits = 0
    int_zero = False

    # String parsing state.
    #
    # While reading STR_LEN, str_val is the parsed string length.
    # While reading STR_BODY, str_val is the number of body characters remaining.
    str_val = 0
    str_len_digits = 0
    str_len_zero = False

    # Whether the top-level object has started/completed.
    root_started = False
    root_completed = False

    # Sum of g_cost(frame) over all open frames.
    g_sum = 0

    # We never need string lengths larger than this cap for feasibility checks.
    # If a length exceeds m, then completion is certainly too long.
    CAP = m + 1

    def g_cost(frame: int) -> int:
        """
        Cost to complete this frame assuming the currently active item inside it
        has just completed.
        """
        if frame == DICT_K:
            # A key just finished; need shortest value "0:" and closing 'e'.
            return 3
        else:
            # List or dictionary after value can close with 'e'.
            return 1

    def f_cost(frame: int) -> int:
        """
        Cost to complete this frame from a clean boundary directly inside it.
        """
        if frame == DICT_V:
            # Must provide shortest value "0:" and then close with 'e'.
            return 3
        else:
            # List or dictionary expecting key can close with 'e'.
            return 1

    def min_remaining() -> int:
        """
        Return the minimum number of additional characters needed to finish the
        current prefix as a valid bencoded object.
        """
        nonlocal progress, root_started, root_completed
        nonlocal frames, g_sum
        nonlocal int_digits, str_val

        # Already complete: need nothing.
        if root_completed:
            return 0

        # At a clean boundary, not inside an integer/string.
        if progress == NONE:
            # If no root item has started, shortest possible object has length 2.
            if not root_started:
                return 2

            # If no frames are open, the root is effectively complete.
            if not frames:
                return 0

            # Complete the innermost frame using f_cost,
            # and all outer frames using their g_cost.
            top = frames[-1]
            return f_cost(top) + (g_sum - g_cost(top))

        # If inside an unfinished integer.
        if progress == INT:
            # Need one digit if no digits yet, then final 'e'.
            finish_cost = (1 if int_digits == 0 else 0) + 1

        # If reading string length.
        elif progress == STR_LEN:
            # Need ':' and then str_val body characters.
            finish_cost = 1 + str_val

        # If reading string body.
        else:
            # Need exactly str_val more raw characters.
            finish_cost = str_val

        # After this item finishes, all open containers still need closing.
        return finish_cost + g_sum

    def item_completed() -> None:
        """
        Called whenever a full item has just been parsed.
        Updates the parent container state, or marks root complete.
        """
        nonlocal root_completed, g_sum

        # If no container is open, this item is the root object.
        if not frames:
            root_completed = True
            return

        # Otherwise, update dictionary state if needed.
        top = frames[-1]

        if top == DICT_K:
            # A dictionary key completed; now it expects a value.
            g_sum -= g_cost(DICT_K)
            frames[-1] = DICT_V
            g_sum += g_cost(DICT_V)

        elif top == DICT_V:
            # A dictionary value completed; now it expects another key or 'e'.
            g_sum -= g_cost(DICT_V)
            frames[-1] = DICT_K
            g_sum += g_cost(DICT_K)

        # If top is LIST, nothing changes.

    def start_item(ch: str) -> bool:
        """
        Try to start a new bencoded item with character ch.
        Return True if successful, False otherwise.
        """
        nonlocal progress, int_digits, int_zero
        nonlocal str_val, str_len_digits, str_len_zero
        nonlocal g_sum

        if ch == "i":
            # Start integer.
            progress = INT
            int_digits = 0
            int_zero = False
            return True

        if ch == "l":
            # Start list.
            frames.append(LIST)
            g_sum += g_cost(LIST)
            return True

        if ch == "d":
            # Start dictionary, initially expecting a key.
            frames.append(DICT_K)
            g_sum += g_cost(DICT_K)
            return True

        if "0" <= ch <= "9":
            # Start string length.
            progress = STR_LEN
            str_val = ord(ch) - ord("0")
            str_len_digits = 1
            str_len_zero = (ch == "0")
            return True

        # No other character can begin an item.
        return False

    def process(ch: str) -> bool:
        """
        Process one input character.
        Return False if the prefix becomes syntactically invalid.
        """
        nonlocal progress
        nonlocal int_digits, int_zero
        nonlocal str_val, str_len_digits, str_len_zero
        nonlocal root_started, root_completed
        nonlocal g_sum

        # Extra characters after the root object are forbidden.
        if root_completed:
            return False

        # Currently reading integer.
        if progress == INT:
            if "0" <= ch <= "9":
                # Leading zero is illegal unless integer is exactly "0".
                if int_zero and int_digits >= 1:
                    return False

                if ch == "0" and int_digits == 0:
                    int_zero = True

                int_digits += 1
                return True

            if ch == "e":
                # Integer must contain at least one digit.
                if int_digits == 0:
                    return False

                progress = NONE
                item_completed()
                return True

            return False

        # Currently reading string length.
        if progress == STR_LEN:
            if "0" <= ch <= "9":
                # Leading zero in string length is illegal unless length is 0.
                if str_len_zero and str_len_digits >= 1:
                    return False

                # Accumulate length, but cap to avoid huge Python integers.
                if str_val <= CAP:
                    str_val = str_val * 10 + (ord(ch) - ord("0"))
                    if str_val > CAP:
                        str_val = CAP

                str_len_digits += 1
                return True

            if ch == ":":
                # Start reading string body.
                progress = STR_BODY

                # Empty string completes immediately.
                if str_val == 0:
                    progress = NONE
                    item_completed()

                return True

            return False

        # Currently reading string body.
        if progress == STR_BODY:
            # Any character is allowed in string body.
            str_val -= 1

            # String completes after exactly the required number of chars.
            if str_val == 0:
                progress = NONE
                item_completed()

            return True

        # progress == NONE: we are at a boundary between items.

        if not frames:
            # If root already started, a new item would be trailing garbage.
            if root_started:
                return False

            # Start root object.
            root_started = True
            return start_item(ch)

        # Inspect current innermost container.
        top = frames[-1]

        if top == LIST or top == DICT_K:
            # Lists and dictionaries expecting keys may close with 'e'.
            if ch == "e":
                g_sum -= g_cost(top)
                frames.pop()

                # The closed list/dictionary is itself one completed item.
                item_completed()
                return True

            # Otherwise, start an item inside the container.
            return start_item(ch)

        # top == DICT_V: dictionary expects a value.
        # It cannot close here.
        if ch == "e":
            return False

        # Start dictionary value.
        return start_item(ch)

    n = len(c)

    # Largest feasible prefix length found so far.
    j_max = 0

    # Scan input.
    for i, ch in enumerate(c):
        # Stop on syntax error.
        if not process(ch):
            break

        # Current prefix length.
        new_j = i + 1

        # Stop if even the shortest completion exceeds m.
        if new_j + min_remaining() > m:
            break

        # This prefix can still be completed within m.
        j_max = new_j

    # Valid iff entire input was consumed as a complete root object.
    if j_max == n and root_completed:
        print("ok")
    else:
        print(f"Error at position {j_max}!")


if __name__ == "__main__":
    solve()
```

---

## 5. Compressed Editorial

Parse the string in one pass with a stack.

Each open container is one of:

- `LIST`
- `DICT_K`: dictionary expecting key or closing `e`
- `DICT_V`: dictionary expecting value

The parser is also in one of:

- `NONE`
- `INT`
- `STR_LEN`
- `STR_BODY`

After each consumed character, check both:

1. syntax is still valid,
2. the prefix can still be completed within total length `m`.

To do the second check in `O(1)`, maintain `g_sum`, the sum of completion costs of all open frames.

For a frame, define:

```text
g_cost(LIST)   = 1
g_cost(DICT_K) = 3
g_cost(DICT_V) = 1
```

This is the cost to finish the frame after the currently parsed item inside it completes.

Also define clean-boundary costs:

```text
f_cost(LIST)   = 1
f_cost(DICT_K) = 1
f_cost(DICT_V) = 3
```

At a clean boundary inside the top frame, the minimum remaining length is:

```text
f_cost(top) + g_sum - g_cost(top)
```

If currently parsing an unfinished atom:

- integer needs either `e`, or digit plus `e`,
- string length needs `:` plus body,
- string body needs remaining body characters,

then add `g_sum`.

Track the largest prefix length `j_max` satisfying:

```text
prefix_length + min_remaining <= m
```

If the whole input is consumed and the root object is complete, print `ok`; otherwise print `Error at position j_max!`.

Complexity:

```text
O(n) time
O(n) memory in worst-case nesting
```
