## 1. Abridged Problem Statement

You are given one line containing a non-empty sequence of valid MML rest commands.

A rest command has the form:

```text
R + duration
```

where duration is one of:

```text
1, 2, 4, 8, 16, 32, 64
```

optionally followed by dots. A dot adds half of the previous duration, but durations smaller than `1/64` are not allowed.

Examples:

```text
R4.  = R4 + R8
R4.. = R4 + R8 + R16
```

Given a sequence of rest commands, output an equivalent sequence with:

1. Minimum possible number of characters.
2. If multiple shortest expressions exist, the lexicographically smallest one.

Input length is at most `100000`.

---

## 2. Detailed Editorial

### Key observation: use units of `1/64`

All possible rest durations are multiples of `1/64`.

So we measure every duration as an integer number of `1/64` units.

For example:

| Command | Duration | Value in `1/64` units |
|---|---:|---:|
| `R1` | `1` | `64` |
| `R2` | `1/2` | `32` |
| `R4` | `1/4` | `16` |
| `R8` | `1/8` | `8` |
| `R16` | `1/16` | `4` |
| `R32` | `1/32` | `2` |
| `R64` | `1/64` | `1` |

Dots add half of the previous value.

For example:

```text
R4.. = R4 + R8 + R16
     = 16 + 8 + 4
     = 28
```

---

### Generate all possible single rest commands

There are only finitely many valid rest commands.

For base `1`, we can have up to 6 dots:

```text
R1, R1., R1.., R1..., R1...., R1....., R1......
```

For base `2`, up to 5 dots.

For base `4`, up to 4 dots.

And so on.

For base `64`, no dots are allowed.

Total number of possible commands:

```text
7 + 6 + 5 + 4 + 3 + 2 + 1 = 28
```

For each command, store:

```text
value  // duration in 1/64 units
len    // string length
str    // actual command string
```

---

### Parse the input

Read the input as one string.

Scan it from left to right.

Each command starts with `'R'`, then has digits, then possibly dots.

For each parsed command, compute its total duration in `1/64` units and add it to the total value `V`.

The problem is now:

> Represent integer `V` as a sum of command values, minimizing total string length. If tied, choose lexicographically smallest concatenation.

---

### Dynamic programming for minimum length

Let:

```text
dp[v] = minimum number of characters needed to represent duration v
```

Base case:

```text
dp[0] = 0
```

Transition:

For every value `v`, try every possible single command `c`.

If `c.val <= v`, then:

```text
dp[v] = min(dp[v], dp[v - c.val] + c.len)
```

This is an unbounded knapsack / shortest path style DP.

The number of commands is only `28`, so this is efficient enough.

---

### Recovering the lexicographically smallest answer

After computing `dp`, we reconstruct the answer greedily.

Suppose we still need value `v`.

A command `c` is feasible if using it first can still lead to an optimal solution:

```text
c.val <= v
dp[v - c.val] + c.len == dp[v]
```

Among feasible commands, we need to choose the one that makes the final string lexicographically smallest.

Naively, one might think we should choose the lexicographically smallest command string. That is almost true, but there is a subtle prefix issue.

For example:

```text
R1
R16
```

Here `R1` is a prefix of `R16`.

If we choose `R1`, the next character after it in the full expression will be another command's `'R'`.

If we choose `R16`, the next character is `'6'`.

Since:

```text
'6' < 'R'
```

the longer command `R16` is lexicographically smaller than starting with `R1...`, assuming both are feasible and preserve optimal length.

Similarly:

```text
R1 <prefix of> R1.
```

After `R1`, the next character would be `'R'`, but in `R1.` the next character is `'.'`.

Since:

```text
'.' < 'R'
```

the longer command wins.

Therefore:

1. Collect all feasible commands.
2. Remove any command that is a strict prefix of another feasible command.
3. Among the remaining commands, choose the lexicographically smallest command string.

Then append it to the answer, subtract its value, and continue.

---

### Correctness sketch

The DP correctly computes the minimum output length because it considers every valid command as the first command of an optimal representation.

During reconstruction, all chosen commands preserve optimal total length.

For lexicographic order:

- If two feasible commands differ before either one ends, the smaller command string gives the smaller full answer.
- If one command is a strict prefix of another, the longer command is lexicographically smaller in the final answer because the shorter command would be followed by `'R'`, while the longer command continues with `'.'` or a digit, both smaller than `'R'`.

Thus the greedy reconstruction produces the lexicographically smallest among all shortest solutions.

---

## 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 commands;

void read() { getline(cin, commands); }

struct Cmd {
    int val;
    int len;
    string str;
};

inline vector<Cmd> generate_commands() {
    vector<Cmd> res;
    int bases[] = {1, 2, 4, 8, 16, 32, 64};
    int unit_vals[] = {64, 32, 16, 8, 4, 2, 1};
    for(int i = 0; i < 7; i++) {
        string s = "R" + to_string(bases[i]);
        int half = unit_vals[i];
        int total = half;
        res.push_back({total, (int)s.size(), s});
        while(half > 1) {
            half /= 2;
            total += half;
            s += '.';
            res.push_back({total, (int)s.size(), s});
        }
    }
    return res;
}

inline int parse_total_value(const string& cmds) {
    int total = 0;
    int n = cmds.size();
    int i = 0;
    while(i < n) {
        i++;
        int base = 0;
        while(i < n && isdigit((unsigned char)cmds[i])) {
            base = base * 10 + (cmds[i] - '0');
            i++;
        }
        int unit = 64 / base;
        int sub = unit;
        while(i < n && cmds[i] == '.') {
            unit /= 2;
            sub += unit;
            i++;
        }
        total += sub;
    }
    return total;
}

inline bool is_strict_prefix(const string& a, const string& b) {
    return a.size() < b.size() && b.compare(0, a.size(), a) == 0;
}

inline vector<int> compute_dp(int V, const vector<Cmd>& cmds) {
    const int INF = INT_MAX;
    vector<int> dp(V + 1, INF);
    dp[0] = 0;
    for(int v = 1; v <= V; v++) {
        for(const auto& c: cmds) {
            if(c.val <= v && dp[v - c.val] != INF) {
                int cand = dp[v - c.val] + c.len;
                if(cand < dp[v]) {
                    dp[v] = cand;
                }
            }
        }
    }
    return dp;
}

inline int pick_best_command(
    int v, const vector<Cmd>& cmds, const vector<int>& dp
) {
    vector<int> feasible;
    for(int i = 0; i < (int)cmds.size(); i++) {
        const auto& c = cmds[i];
        if(c.val <= v && dp[v - c.val] != INT_MAX &&
           dp[v - c.val] + c.len == dp[v]) {
            feasible.push_back(i);
        }
    }

    vector<int> survivors;
    for(int i: feasible) {
        bool dominated = false;
        for(int j: feasible) {
            if(i != j && is_strict_prefix(cmds[i].str, cmds[j].str)) {
                dominated = true;
                break;
            }
        }
        if(!dominated) {
            survivors.push_back(i);
        }
    }

    int best = survivors[0];
    for(int i: survivors) {
        if(cmds[i].str < cmds[best].str) {
            best = i;
        }
    }
    return best;
}

void solve() {
    // Work in 1/64 units so every duration becomes a positive integer.
    // Each rest command is "R" + base + k dots, with base in
    // {1,2,4,8,16,32,64} and the dotted half-chain not falling below the
    // 1/64 unit; this gives 7+6+5+4+3+2+1 = 28 distinct commands. Let V be
    // the total value of the input. Then dp[v] = shortest expression for
    // value v is an unbounded-knapsack DP over the 28 commands.
    //
    // For lex-smallest recovery, observe that every character that may
    // appear inside a command past the leading 'R' is either a digit
    // ('0'..'9' = 48..57) or '.' (46), all strictly less than 'R' (82).
    // After a command ends, the next character is the start of the next
    // command, which is always 'R'. So if at value v two feasible
    // candidates a, b satisfy "a.str is a strict prefix of b.str", their
    // full answers agree on the first |a.str| chars; at index |a.str|,
    // candidate a's char comes from the next command (it's 'R') while
    // candidate b's char is a digit or '.', strictly smaller. Hence b
    // always beats a, and we can discard every feasible command dominated
    // by a longer feasible command sharing its prefix. The survivors then
    // have pairwise non-prefix-related strings, so their full answers
    // differ inside their own c.str and the lex-smallest c.str wins.
    // Recovery is therefore a plain string compare among at most 28
    // fixed strings at each step.

    auto cmds = generate_commands();
    int V = parse_total_value(commands);
    auto dp = compute_dp(V, cmds);

    string ans;
    int v = V;
    while(v > 0) {
        int idx = pick_best_command(v, cmds, dp);
        ans += cmds[idx].str;
        v -= cmds[idx].val;
    }
    cout << ans << "\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


class Cmd:
    """
    Represents one valid rest command.
    """

    def __init__(self, val, length, string):
        # Duration in units of 1/64.
        self.val = val

        # Number of characters in this command.
        self.len = length

        # Actual command string, e.g. "R4..".
        self.str = string


def generate_commands():
    """
    Generate all valid single rest commands.

    There are 28 commands total:
    R1 with 0..6 dots,
    R2 with 0..5 dots,
    ...
    R64 with 0 dots.
    """

    # Possible base denominators.
    bases = [1, 2, 4, 8, 16, 32, 64]

    # Corresponding durations in 1/64 units.
    unit_vals = [64, 32, 16, 8, 4, 2, 1]

    cmds = []

    # Generate every base and all legal dotted versions.
    for base, unit_value in zip(bases, unit_vals):
        # Command without dots.
        s = "R" + str(base)

        # Current added duration part.
        half = unit_value

        # Total duration of this command.
        total = half

        # Add command without dots.
        cmds.append(Cmd(total, len(s), s))

        # Add dotted versions while the next dot still contributes at least 1/64.
        while half > 1:
            # Dot contributes half of the previous contribution.
            half //= 2

            # Increase command duration.
            total += half

            # Add dot to string.
            s += "."

            # Store this dotted command.
            cmds.append(Cmd(total, len(s), s))

    return cmds


def parse_total_value(commands):
    """
    Parse the input string and compute total duration in 1/64 units.
    """

    n = len(commands)
    i = 0
    total = 0

    # Parse command by command.
    while i < n:
        # Current character is 'R'.
        i += 1

        # Parse duration base number.
        base = 0
        while i < n and commands[i].isdigit():
            base = base * 10 + int(commands[i])
            i += 1

        # Convert base duration to 1/64 units.
        unit = 64 // base

        # Command duration starts with base unit.
        subtotal = unit

        # Parse dots.
        while i < n and commands[i] == ".":
            # Each dot adds half of the previous part.
            unit //= 2

            # Add this dot's contribution.
            subtotal += unit

            # Move past dot.
            i += 1

        # Add command duration to total.
        total += subtotal

    return total


def is_strict_prefix(a, b):
    """
    Return True if a is a strict prefix of b.
    """

    return len(a) < len(b) and b.startswith(a)


def compute_dp(V, cmds):
    """
    Compute dp[v] = minimum number of characters needed
    to represent duration v.
    """

    INF = 10**18

    # dp[v] is the shortest length for duration v.
    dp = [INF] * (V + 1)

    # Duration zero is represented by the empty string.
    dp[0] = 0

    # Compute in increasing order.
    for v in range(1, V + 1):
        best = INF

        # Try every command as the last command.
        for c in cmds:
            if c.val <= v:
                prev = dp[v - c.val]

                # If previous duration is reachable, update best.
                if prev != INF:
                    cand = prev + c.len
                    if cand < best:
                        best = cand

        dp[v] = best

    return dp


def pick_best_command(v, cmds, dp):
    """
    Choose the lexicographically best first command among all commands
    that can start an optimal solution for duration v.
    """

    feasible = []

    # Find all commands preserving the optimal length.
    for idx, c in enumerate(cmds):
        if c.val <= v and dp[v - c.val] + c.len == dp[v]:
            feasible.append(idx)

    survivors = []

    # Remove commands that are strict prefixes of another feasible command.
    for i in feasible:
        dominated = False

        for j in feasible:
            if i != j and is_strict_prefix(cmds[i].str, cmds[j].str):
                dominated = True
                break

        if not dominated:
            survivors.append(i)

    # Among survivors, ordinary string comparison decides.
    best = survivors[0]

    for i in survivors:
        if cmds[i].str < cmds[best].str:
            best = i

    return best


def solve():
    # Read the single input line.
    commands = sys.stdin.readline().strip()

    # Generate all possible rest commands.
    cmds = generate_commands()

    # Compute total input duration.
    V = parse_total_value(commands)

    # Compute shortest lengths.
    dp = compute_dp(V, cmds)

    # Reconstruct lexicographically smallest shortest answer.
    ans = []
    v = V

    while v > 0:
        # Pick the best next command.
        idx = pick_best_command(v, cmds, dp)

        # Append command string.
        ans.append(cmds[idx].str)

        # Subtract command duration.
        v -= cmds[idx].val

    # Print final expression.
    print("".join(ans))


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

---

## 5. Compressed Editorial

Convert all durations to integer units of `1/64`.

Generate all valid single rest commands. There are only `28` of them. Store each command's value, length, and string.

Parse the input sequence and compute its total value `V`.

Use unbounded knapsack DP:

```text
dp[0] = 0
dp[v] = min(dp[v - cmd.value] + cmd.length)
```

over all commands fitting into `v`.

This gives the minimum possible output length.

To reconstruct the lexicographically smallest shortest string, repeatedly choose the first command among those satisfying:

```text
dp[v - cmd.value] + cmd.length == dp[v]
```

If one feasible command is a strict prefix of another feasible command, discard the shorter one, because after the shorter command the next character would be `'R'`, while the longer command continues with `'.'` or a digit, both lexicographically smaller than `'R'`.

Among remaining feasible commands, choose the lexicographically smallest string, append it, subtract its value, and continue.

This produces the shortest expression, and among shortest expressions, the lexicographically smallest one.
