<|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 `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: 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: skip `R`, parse the base duration number, convert it to units of `1/64`, add dot contributions, add the result to total duration `V`.

---

### Step 3: Dynamic programming

Compute `dp[v] = shortest output length for total duration v` for all `0 <= v <= V`.

Transition: for each command c, if c.value <= v, then 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 `c.value <= v` and `dp[v - c.value] + c.length == dp[v]`.

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

- If one feasible command is a strict prefix of another feasible command, the longer one 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;

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;
}
```

---

## 5. Python Implementation with Detailed Comments

```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
        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()
```
