## 1. Abridged Problem Statement

Diputs notation represents non-negative integers using 9 digit characters, from least to most significant:

```text
_ . , - ~ = ' ^ "
```

Each digit may appear only a limited number of times:

```text
_ : 2
. : 3
, : 5
- : 7
~ : 11
= : 13
' : 17
^ : 19
" : 23
```

A Diputs number is written from most significant digit to least significant digit, with repeated characters grouped together. Zero is represented by `O`.

The input is arbitrary text containing:

- decimal numbers,
- Diputs numbers,
- other characters.

You must replace every decimal number with its Diputs representation, and every Diputs number with its decimal representation.

Parsing must be greedy: always take the longest valid decimal or Diputs number possible. Decimal numbers cannot have leading zeroes, so `020` is parsed as `0` and `20`.

If the total number of parsed numbers is odd, sort the numeric values increasingly before output, while preserving the original positions' output notation types.

Input and output sizes are at most `10^6` bytes.

---

## 2. Detailed Editorial

### Diputs as a mixed-radix numeral system

The Diputs digits are:

```text
index:  0 1 2 3 4 5 6 7 8
char:   _ . , - ~ = ' ^ "
```

The maximum count of each digit is:

```text
2, 3, 5, 7, 11, 13, 17, 19, 23
```

For digit `i`, the number of possible counts is:

```text
max_count[i] + 1
```

For example, `_` can appear `0`, `1`, or `2` times, so it has base `3`.

The ordering rule says that numbers are ordered first by the count of the most significant digit `"`, then by `^`, then by `'`, and so on.

This is equivalent to a mixed-radix system.

Define:

```text
product[0] = 1
product[i + 1] = product[i] * (max_count[i] + 1)
```

Then the decimal value of a Diputs number is:

```text
value = count[0] * product[0]
      + count[1] * product[1]
      + ...
      + count[8] * product[8]
```

The maximum representable number is:

```text
product[9] - 1
```

which equals:

```text
836075519
```

Zero is special and is represented by `O`.

---

### Converting decimal to Diputs

Given a decimal number `n`.

If `n == 0`, output:

```text
O
```

Otherwise, compute each digit count:

```text
count[i] = (n / product[i]) % (max_count[i] + 1)
```

Then output the digits from most significant to least significant:

```text
" repeated count[8] times
^ repeated count[7] times
...
_ repeated count[0] times
```

---

### Converting Diputs to decimal

For a Diputs token, every character directly contributes the value of its digit.

For example:

```text
,..__
```

means:

```text
1 comma + 2 dots + 2 underscores
```

So its decimal value is:

```text
product[2] + 2 * product[1] + 2 * product[0]
```

The implementation simply iterates through the token and adds:

```text
product[index_of_character]
```

for each character.

---

### Greedy parsing

The input is arbitrary text. We scan from left to right.

At each position:

#### 1. Decimal number

If the current character is a digit:

- If it is `0`, parse exactly one character as number `0`.
- Otherwise, keep taking digits while the resulting decimal number does not exceed the maximum Diputs value.

This handles cases like:

```text
020
```

as:

```text
0 and 20
```

and very large digit strings are split greedily.

---

#### 2. Diputs zero

If the current character is `O`, parse it as Diputs zero.

---

#### 3. Nonzero Diputs number

If the current character is one of:

```text
_ . , - ~ = ' ^ "
```

parse the longest valid Diputs representation.

A valid Diputs representation must:

1. Be written from most significant to least significant.
2. Not exceed the maximum count for any digit.

Using digit indices from least significant to most significant, this means indices must be non-increasing while reading left to right.

For example:

```text
,..__
```

has indices:

```text
2, 1, 1, 0, 0
```

which is valid.

But:

```text
_.
```

has indices:

```text
0, 1
```

which is invalid as one Diputs number, so it is parsed as:

```text
_
.
```

Also:

```text
___
```

cannot be one number because `_` may appear at most twice, so it is parsed as:

```text
__
_
```

---

#### 4. Other characters

Any other character is copied unchanged and acts as a separator.

---

### Sorting rule

After tokenization, suppose we found `k` numbers.

Each token stores:

- its position in the input,
- its original length,
- its numeric value,
- whether it was originally decimal.

Normally, each token is converted to the opposite notation independently.

If `k` is odd:

1. Extract all token values.
2. Sort them.
3. Assign sorted values back to the token positions in order.

Important: the notation type of each position is preserved.

That is, if a position originally contained a decimal number, then after sorting it still outputs Diputs notation. If a position originally contained a Diputs number, then after sorting it still outputs decimal notation.

---

### Complexity

Let `N` be the input size.

Tokenization is linear:

```text
O(N)
```

Sorting happens only on the parsed numbers:

```text
O(K log K)
```

where `K <= N`.

Memory usage is:

```text
O(N)
```

---

## 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;
};

string input_text;

void read() {
    stringstream buffer;
    buffer << cin.rdbuf();
    input_text = buffer.str();
}

void solve() {
    // Pure implementation. Tokenize the input into decimal numbers, Diputs
    // numbers and "other" characters. A Diputs digit is one of `_.,-~='^"`
    // (least to most significant); a valid Diputs representation lists digits
    // in non-increasing significance and respects per-digit max quantities, so
    // we parse greedily under both constraints. Decimal numbers are parsed
    // greedily up to the Diputs maximum value, with a leading '0' forming the
    // single number 0. The encoding is a mixed-radix system, so conversion in
    // either direction is a small loop instead of enumerating ~836M values.
    // After tokenizing we flip each token's notation, optionally sort the
    // values when the total count is odd, and emit with original separators.

    const string diputs_chars = "_.,-~='^\"";
    const int max_counts[9] = {2, 3, 5, 7, 11, 13, 17, 19, 23};

    int64_t products[10];
    products[0] = 1;
    for(int i = 0; i < 9; i++) {
        products[i + 1] = products[i] * (max_counts[i] + 1);
    }

    int64_t max_num = products[9] - 1;

    array<int, 256> sig_of;
    sig_of.fill(-1);
    for(int i = 0; i < 9; i++) {
        sig_of[(unsigned char)diputs_chars[i]] = i;
    }

    auto diputs_to_decimal = [&](size_t start, size_t end) -> int64_t {
        int64_t value = 0;
        for(size_t k = start; k < end; k++) {
            value += products[sig_of[(unsigned char)input_text[k]]];
        }
        return value;
    };

    auto decimal_to_diputs = [&](int64_t n) -> string {
        if(n == 0) {
            return "O";
        }
        int counts[9];
        for(int i = 0; i < 9; i++) {
            counts[i] = (n / products[i]) % (max_counts[i] + 1);
        }
        string result;
        for(int i = 8; i >= 0; i--) {
            result.append(counts[i], diputs_chars[i]);
        }
        return result;
    };

    struct token {
        size_t start;
        size_t length;
        int64_t value;
        bool was_decimal;
    };

    vector<token> tokens;

    size_t n = input_text.size();
    for(size_t i = 0; i < n;) {
        unsigned char c = input_text[i];
        if(isdigit(c)) {
            size_t j = i;
            int64_t value = 0;
            if(c == '0') {
                j = i + 1;
            } else {
                while(j < n && isdigit((unsigned char)input_text[j])) {
                    int64_t next_val = value * 10 + (input_text[j] - '0');
                    if(next_val > max_num) {
                        break;
                    }
                    value = next_val;
                    j++;
                }
            }
            tokens.push_back({i, j - i, value, true});
            i = j;
        } else if(c == 'O') {
            tokens.push_back({i, 1, 0, false});
            i++;
        } else if(sig_of[c] >= 0) {
            size_t j = i;
            int last_sig = 8;
            int counts[9] = {};
            while(j < n) {
                int s = sig_of[(unsigned char)input_text[j]];
                if(s < 0 || s > last_sig || counts[s] >= max_counts[s]) {
                    break;
                }
                counts[s]++;
                last_sig = s;
                j++;
            }
            tokens.push_back({i, j - i, diputs_to_decimal(i, j), false});
            i = j;
        } else {
            i++;
        }
    }

    if(tokens.size() % 2 == 1) {
        vector<int64_t> values;
        values.reserve(tokens.size());
        for(auto& t: tokens) {
            values.push_back(t.value);
        }
        sort(values.begin(), values.end());
        for(size_t k = 0; k < tokens.size(); k++) {
            tokens[k].value = values[k];
        }
    }

    string result;
    result.reserve(input_text.size() * 2);
    size_t cursor = 0;
    for(auto& t: tokens) {
        while(cursor < t.start) {
            result += input_text[cursor++];
        }
        if(t.was_decimal) {
            result += decimal_to_diputs(t.value);
        } else {
            result += to_string(t.value);
        }
        cursor = t.start + t.length;
    }

    while(cursor < input_text.size()) {
        result += input_text[cursor++];
    }

    cout << result;
}

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 with Detailed Comments

```python
import sys


def main():
    # Read the entire input as one string.
    text = sys.stdin.read()

    # Diputs digits from least significant to most significant.
    diputs_chars = "_.,-~='^\""

    # Maximum allowed counts for each Diputs digit.
    max_counts = [2, 3, 5, 7, 11, 13, 17, 19, 23]

    # products[i] is the mixed-radix place value of digit i.
    products = [1]

    # Build place values.
    # products[i + 1] = products[i] * number_of_possible_counts_for_digit_i
    for cnt in max_counts:
        products.append(products[-1] * (cnt + 1))

    # Maximum representable value.
    max_num = products[9] - 1

    # Map each Diputs character to its significance index.
    sig_of = {ch: i for i, ch in enumerate(diputs_chars)}

    def diputs_to_decimal(s):
        """
        Convert a valid nonzero Diputs representation to decimal.
        """
        value = 0

        # Each character contributes its place value.
        for ch in s:
            value += products[sig_of[ch]]

        return value

    def decimal_to_diputs(n):
        """
        Convert a decimal integer to Diputs notation.
        """
        # Zero is special.
        if n == 0:
            return "O"

        counts = [0] * 9

        # Extract mixed-radix digits.
        for i in range(9):
            counts[i] = (n // products[i]) % (max_counts[i] + 1)

        result = []

        # Output from most significant digit to least significant digit.
        for i in range(8, -1, -1):
            if counts[i]:
                result.append(diputs_chars[i] * counts[i])

        return "".join(result)

    # Each token is represented as:
    # [start_position, length, numeric_value, was_decimal]
    tokens = []

    n = len(text)
    i = 0

    # Greedily scan input from left to right.
    while i < n:
        c = text[i]

        # Case 1: decimal number.
        if c.isdigit():
            j = i
            value = 0

            # Decimal numbers may not have leading zeroes.
            # Therefore '0' is always a token of length 1.
            if c == "0":
                j = i + 1
            else:
                # Extend while the token remains a valid decimal number
                # not exceeding max_num.
                while j < n and text[j].isdigit():
                    next_value = value * 10 + (ord(text[j]) - ord("0"))

                    # Stop before exceeding the representable range.
                    if next_value > max_num:
                        break

                    value = next_value
                    j += 1

            # Store decimal token.
            tokens.append([i, j - i, value, True])

            # Continue after token.
            i = j

        # Case 2: Diputs zero.
        elif c == "O":
            tokens.append([i, 1, 0, False])
            i += 1

        # Case 3: nonzero Diputs number.
        elif c in sig_of:
            j = i

            # Previous significance index.
            # A valid Diputs representation must have non-increasing
            # significance from left to right.
            last_sig = 8

            # Counts used in this token.
            counts = [0] * 9

            # Greedily extend the Diputs token.
            while j < n:
                ch = text[j]

                # Stop if not a Diputs digit.
                if ch not in sig_of:
                    break

                s = sig_of[ch]

                # Stop if significance increases.
                if s > last_sig:
                    break

                # Stop if this digit has reached its maximum allowed count.
                if counts[s] >= max_counts[s]:
                    break

                # Accept this character.
                counts[s] += 1

                # Future characters must be no more significant.
                last_sig = s

                # Advance end position.
                j += 1

            # Convert the parsed substring to decimal.
            value = diputs_to_decimal(text[i:j])

            # Store Diputs token.
            tokens.append([i, j - i, value, False])

            # Continue after token.
            i = j

        # Case 4: ordinary non-number text.
        else:
            i += 1

    # If there is an odd number of numeric tokens, sort their values.
    if len(tokens) % 2 == 1:
        values = sorted(token[2] for token in tokens)

        # Put sorted values back into token positions.
        # The original notation type at each position is preserved.
        for idx, value in enumerate(values):
            tokens[idx][2] = value

    # Build the final output.
    result = []
    cursor = 0

    for start, length, value, was_decimal in tokens:
        # Copy unchanged text before this token.
        if cursor < start:
            result.append(text[cursor:start])

        # Convert token to the opposite notation.
        if was_decimal:
            result.append(decimal_to_diputs(value))
        else:
            result.append(str(value))

        # Skip original token.
        cursor = start + length

    # Copy remaining text after the last token.
    if cursor < n:
        result.append(text[cursor:])

    # Write final answer.
    sys.stdout.write("".join(result))


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

---

## 5. Compressed Editorial

Treat Diputs notation as a mixed-radix system.

For digit counts:

```text
max = [2, 3, 5, 7, 11, 13, 17, 19, 23]
```

define:

```text
prod[0] = 1
prod[i + 1] = prod[i] * (max[i] + 1)
```

Then a Diputs number has value:

```text
sum(count[i] * prod[i])
```

and decimal-to-Diputs is done by:

```text
count[i] = (n // prod[i]) % (max[i] + 1)
```

Zero is represented by `O`.

Scan the entire input greedily:

- If current character is a decimal digit:
  - parse `0` as a single digit,
  - otherwise consume digits while value does not exceed the maximum representable number.
- If current character is `O`, parse Diputs zero.
- If current character is a Diputs digit, consume the longest valid Diputs token:
  - digit significances must be non-increasing,
  - no digit count may exceed its maximum.
- Otherwise skip it as normal text.

Store each token's position, length, numeric value, and original notation type.

If the number of tokens is odd, sort only their numeric values and assign them back to token positions. The notation type of each position stays unchanged.

Finally reconstruct the text, copying separators unchanged and converting each token to the opposite notation.

Complexity:

```text
O(N + K log K)
```

where `N` is input size and `K` is the number of parsed numbers.
