<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

447. Optimal Rest
Time limit per test: 1 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Music Macro Language (MML) is a language for textual representation of musical scores. Although there are various dialects of MML, all of them provide a set of commands to describe scores, such as commands for notes, rests, octaves, volumes, and so forth. In this problem, we focus on rests, i.e. intervals of silence. Each rest command consists of a command specifier 'R' followed by a duration specifier. Each duration specifier is basically one of the following numbers: '1', '2', '4', '8', '16', '32', and '64', where '1' denotes a whole (1), '2' a half (1/2), '4' a quarter (1/4), '8' an eighth (1/8), and so on. This number is called the base duration, and optionally followed by one or more dots. The first dot adds the duration by the half of the base duration. For example, '4.' denotes the duration of '4' (a quarter) plus '8' (an eighth, i.e. the half of a quarter), or simply 1.5 times as long as '4'. In other words, 'R4.' is equivalent to 'R4R8'. In case with two or more dots, each extra dot extends the duration by the half of the previous one. Thus '4..' denotes the duration of '4' plus '8' plus '16', '4...' denotes the duration of '4' plus '8' plus '16' plus '32', and so on. The duration extended by dots cannot be shorter than '64'. For exapmle, neither '64.' nor '16...' will be accepted since both of the last dots indicate the half of '64' (i.e. the duration of 1/128). In this problem, you are required to write a program that finds the shortest expressions equivalent to given sequences of rest commands.
Input
The input consists of a line containing a non-empty sequence of valid rest commands. You may assume that the sequence does not contain more than 100000 characters.
Output
Your program should output the shortest expression in one line. If there are multiple expressions of the shortest length, output the lexicographically smallest one.
Example(s)
sample input
sample output
R2R2
R1

sample input
sample output
R1R2R4R8R16R32R64
R1......

sample input
sample output
R1R4R16
R16R1R4

<|response|>
## 1. Abridged Problem Statement

You are given a non-empty sequence of MML rest commands.

Each command has form:

```text
R + base duration + optional dots
```

where base duration is one of:

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

A dot adds half of the previous duration, but no added part may be shorter than `1/64`.

Given the sequence, output an equivalent rest sequence with:

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

The input length is at most `100000`.

---

## 2. Key Observations

### Observation 1: Only total duration matters

All commands are rests, i.e. silence. Therefore, two sequences are equivalent if their total durations are equal.

So we do **not** need to preserve command order.

Example:

```text
R1R4R16
```

has the same duration as:

```text
R16R1R4
```

and the latter may be lexicographically smaller.

---

### Observation 2: Use units of `1/64`

Every valid rest duration is a multiple of `1/64`.

Represent durations as integers:

| Command | Duration | Integer value |
|---|---:|---:|
| `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` |

For example:

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

---

### Observation 3: There are only 28 possible single commands

For each base duration, the number of valid dots is limited.

```text
R1  can have 0..6 dots
R2  can have 0..5 dots
R4  can have 0..4 dots
...
R64 can have 0 dots
```

Total number of possible commands:

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

So we can generate all possible single rest commands.

---

### Observation 4: Shortest output is an unbounded knapsack DP

Let:

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

Then:

```text
dp[0] = 0
dp[v] = min(dp[v - command_value] + command_length)
```

over all valid commands.

---

### Observation 5: Lexicographically smallest reconstruction needs care

Suppose two feasible first commands are:

```text
R1
R16
```

Normally, `R1` is a prefix of `R16`.

But if we output `R1`, the next character after it will be another command's `R`.

Compare:

```text
R1R...
R16...
```

At the third character:

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

So `R16...` is lexicographically smaller.

Similarly:

```text
R1.
```

is lexicographically smaller than:

```text
R1R...
```

because:

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

Therefore, when one feasible command is a strict prefix of another feasible command, the longer command is lexicographically better.

---

## 3. Full Solution Approach

### Step 1: Generate all valid commands

For every base duration:

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

generate the command without dots and then keep adding dots while the next dot contributes at least `1/64`.

For each command store:

```text
value  - duration in units of 1/64
length - number of characters
string - actual command string
```

---

### Step 2: Parse the input

Scan the input string command by command.

For each command:

1. Skip `R`.
2. Parse the base duration number.
3. Convert it to units of `1/64`.
4. Add dot contributions.
5. Add the result to total duration `V`.

---

### Step 3: Dynamic programming

Compute:

```text
dp[v] = shortest output length for total duration v
```

for all `0 <= v <= V`.

Transition:

```text
for each command c:
    if c.value <= v:
        dp[v] = min(dp[v], dp[v - c.value] + c.length)
```

---

### Step 4: Reconstruct the answer greedily

At remaining duration `v`, a command `c` is feasible if:

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

Among all feasible commands, choose the lexicographically best first command.

Comparison rule between two command strings `a` and `b`:

- If `a` is a strict prefix of `b`, then `b` is better.
- If `b` is a strict prefix of `a`, then `a` is better.
- Otherwise, use normal lexicographical comparison.

Append the chosen command, subtract its value, and continue.

---

### Complexity

Let `V` be the total duration in units of `1/64`.

The input length is at most `100000`, so `V` is at most about `3.2 * 10^6`.

There are only `28` commands.

```text
Time complexity:  O(28 * V)
Memory complexity: O(V)
```

---

## 4. C++ Implementation with Detailed Comments

```cpp
#include <bits/stdc++.h>
using namespace std;

struct Command {
    int value;      // Duration measured in units of 1/64
    int length;     // Number of characters in this command
    string text;    // Actual command string, for example "R4.."
};

// Check whether a is a strict prefix of b.
bool isStrictPrefix(const string& a, const string& b) {
    return a.size() < b.size() && b.compare(0, a.size(), a) == 0;
}

// Compare two commands as possible first commands of an optimal answer.
//
// Normally, lexicographical order is enough.
// However, if one command is a prefix of the other, the longer command wins.
// Example:
//   R1R...
//   R16...
// Since '6' < 'R', R16... is lexicographically smaller.
bool betterFirstCommand(const Command& a, const Command& b) {
    if (isStrictPrefix(a.text, b.text)) {
        return false; // b is better
    }

    if (isStrictPrefix(b.text, a.text)) {
        return true; // a is better
    }

    return a.text < b.text;
}

// Generate all 28 valid rest commands.
vector<Command> generateCommands() {
    vector<Command> commands;

    vector<int> bases = {1, 2, 4, 8, 16, 32, 64};

    for (int base : bases) {
        // Base duration in units of 1/64.
        // Example: base = 4 means duration 1/4 = 16/64.
        int part = 64 / base;

        int total = part;
        string text = "R" + to_string(base);

        // Command without dots.
        commands.push_back({total, (int)text.size(), text});

        // Add dots while the next added part is at least 1/64.
        while (part > 1) {
            part /= 2;
            total += part;
            text += '.';

            commands.push_back({total, (int)text.size(), text});
        }
    }

    return commands;
}

// Parse the input sequence and return total duration in units of 1/64.
int parseTotalValue(const string& s) {
    int n = (int)s.size();
    int i = 0;
    int total = 0;

    while (i < n) {
        // Current character is 'R'.
        i++;

        // Parse the base number after 'R'.
        int base = 0;
        while (i < n && isdigit((unsigned char)s[i])) {
            base = base * 10 + (s[i] - '0');
            i++;
        }

        // Convert base duration to units of 1/64.
        int part = 64 / base;
        int subtotal = part;

        // Parse dots.
        while (i < n && s[i] == '.') {
            part /= 2;
            subtotal += part;
            i++;
        }

        total += subtotal;
    }

    return total;
}

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

    string input;
    getline(cin, input);

    vector<Command> commands = generateCommands();

    int V = parseTotalValue(input);

    const int INF = 1e9;

    // dp[v] = minimum number of characters needed for duration v.
    vector<int> dp(V + 1, INF);
    dp[0] = 0;

    // Unbounded knapsack DP.
    for (int v = 1; v <= V; v++) {
        for (const Command& cmd : commands) {
            if (cmd.value <= v) {
                dp[v] = min(dp[v], dp[v - cmd.value] + cmd.length);
            }
        }
    }

    // Precompute command order for lexicographically best reconstruction.
    vector<int> order(commands.size());
    iota(order.begin(), order.end(), 0);

    sort(order.begin(), order.end(), [&](int i, int j) {
        return betterFirstCommand(commands[i], commands[j]);
    });

    string answer;
    answer.reserve(dp[V]);

    int remaining = V;

    // Greedily reconstruct the lexicographically smallest shortest answer.
    while (remaining > 0) {
        for (int idx : order) {
            const Command& cmd = commands[idx];

            // This command can be the first command of an optimal answer.
            if (cmd.value <= remaining &&
                dp[remaining - cmd.value] + cmd.length == dp[remaining]) {
                
                answer += cmd.text;
                remaining -= cmd.value;
                break;
            }
        }
    }

    cout << answer << '\n';

    return 0;
}
```

---

## 5. Python Implementation with Detailed Comments

```python
import sys
from functools import cmp_to_key


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

    __slots__ = ("value", "length", "text")

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

        # Number of characters in the command.
        self.length = length

        # Actual command string, for example "R4..".
        self.text = text


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

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


def compare_first_command(a, b):
    """
    Comparator for commands when used as the first command of the remaining answer.

    If one command is a prefix of the other, the longer one is better.

    Example:
        R1R...
        R16...

    At the third character, '6' < 'R', so R16... is lexicographically smaller.
    """

    if is_strict_prefix(a.text, b.text):
        return 1      # b should come before a

    if is_strict_prefix(b.text, a.text):
        return -1     # a should come before b

    if a.text < b.text:
        return -1
    if a.text > b.text:
        return 1
    return 0


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
    """

    commands = []
    bases = [1, 2, 4, 8, 16, 32, 64]

    for base in bases:
        # Base duration in units of 1/64.
        part = 64 // base

        total = part
        text = "R" + str(base)

        # Command without dots.
        commands.append(Command(total, len(text), text))

        # Add dotted versions while the next dot contributes at least 1/64.
        while part > 1:
            part //= 2
            total += part
            text += "."

            commands.append(Command(total, len(text), text))

    return commands


def parse_total_value(s):
    """
    Parse the input command sequence and compute its total duration
    in units of 1/64.
    """

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

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

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

        # Convert base duration to units of 1/64.
        part = 64 // base
        subtotal = part

        # Parse dots.
        while i < n and s[i] == ".":
            part //= 2
            subtotal += part
            i += 1

        total += subtotal

    return total


def solve():
    input_string = sys.stdin.readline().strip()

    commands = generate_commands()

    V = parse_total_value(input_string)

    INF = 10**9

    # dp[v] = minimum number of characters needed for duration v.
    dp = [INF] * (V + 1)
    dp[0] = 0

    # Sort commands by value only for a tiny DP optimization:
    # once command.value > v, we can stop checking.
    dp_commands = sorted(commands, key=lambda c: c.value)

    # Unbounded knapsack DP.
    for v in range(1, V + 1):
        best = INF

        for cmd in dp_commands:
            if cmd.value > v:
                break

            candidate = dp[v - cmd.value] + cmd.length

            if candidate < best:
                best = candidate

        dp[v] = best

    # Command order for lexicographically smallest reconstruction.
    lex_order = sorted(commands, key=cmp_to_key(compare_first_command))

    answer_parts = []
    remaining = V

    # Greedily reconstruct the lexicographically smallest shortest expression.
    while remaining > 0:
        for cmd in lex_order:
            # The command is feasible if it preserves optimal total length.
            if (
                cmd.value <= remaining
                and dp[remaining - cmd.value] + cmd.length == dp[remaining]
            ):
                answer_parts.append(cmd.text)
                remaining -= cmd.value
                break

    print("".join(answer_parts))


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