<|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.

202. The Towers of Hanoi Revisited
time limit per test: 0.25 sec.
memory limit per test: 65536 KB
input: standard
output: standard



You all must know the puzzle named ``The Towers of Hanoi''. The puzzle has three pegs and N discs of different radii, initially all disks are located on the first peg, ordered by their radii - the largest at the bottom, the mallest at the top. In a turn you may take the topmost disc from any peg and move it to another peg, the only rule says that you may not place the disc atop any smaller disk. The problem is to move all disks to the last peg making the smallest possible number of moves.

There is the legend that somewhere in Tibet there is a monastery where monks tirelessly move disks from peg to peg solving the puzzle for 64 discs. The legend says that when they finish, the end of the world would come.

Since it is well known that to solve the puzzle you need to make 2N-1 moves, a small calculation shows that the world seems to be a quite safe place for a while.

However, recent archeologists discoveries have shown that the things can be a bit worse. The manuscript found in Tibet mountains says that the puzzle the monks are solving has not 3 but M pegs. This is the problem, because when increasing the number of pegs, the number of moves needed to move all discs from the first peg to the last one following the rules described, decreases dramatically.

Calculate how many moves one needs to move N discs from the first peg to the last one when the puzzle has M pegs and provide the scenario for moving the discs.

Input
Input file contains N and M (1 ≤ N ≤ 64, 4 ≤ M ≤ 65).

Output
On the first line output L - the number of moves needed to solve the puzzle. Next L lines must contain the moves themselves. For each move print the line of the form

move <disc-radius> from <source-peg> to <target-peg>

if the disc is moved to the empty peg or

move <disc-radius> from <source-peg> to <target-peg> atop <target-top-disc-radius>

if the disc is moved atop some other disc.

Disc radii are integer numbers from 1 to N, pegs are numbered from 1 to M.

Sample test(s)

Input
5 4

Output
13
move 1 from 1 to 3
move 2 from 1 to 2
move 1 from 3 to 2 atop 2
move 3 from 1 to 4
move 4 from 1 to 3
move 3 from 4 to 3 atop 4
move 5 from 1 to 4
move 3 from 3 to 1
move 4 from 3 to 4 atop 5
move 3 from 1 to 4 atop 4
move 1 from 2 to 1
move 2 from 2 to 4 atop 3
move 1 from 1 to 4 atop 2
Author:	Andrew Stankevich
Resource:	Petrozavodsk Summer Trainings 2003
Date:	2003-08-23

<|response|>
## 1) Abridged problem statement (concise)

There are **N** disks (sizes `1..N`, `1` smallest) initially stacked on **peg 1**. There are **M pegs** (`4 ≤ M ≤ 65`). A move takes the **top** disk from one peg to another, never placing a larger disk on a smaller one.

You must:
1. Output **L**, the number of moves produced by your algorithm.
2. Output a valid sequence of **L** moves to transfer all disks from **peg 1** to **peg M**.
   - Print `move d from a to b` if peg `b` is empty,
   - else `move d from a to b atop x` where `x` is the disk currently on top of peg `b`.

Constraints: `1 ≤ N ≤ 64`.

---

## 2) Key observations

1. **Classic 3-peg formula doesn’t apply.** With more pegs, the minimum number of moves can be much smaller than `2^N - 1`.

2. A standard (and intended) strategy is the **Frame–Stewart recurrence**:
   - To move `n` disks using `m` pegs:
     - Pick `k` (`1 ≤ k < n`) to temporarily move the top `k` disks to some auxiliary peg (using all `m` pegs),
     - Move the remaining `n-k` disks to the target using only `m-1` pegs (because the auxiliary peg is occupied by the `k`-stack),
     - Move the `k` disks from auxiliary onto the target (again using `m` pegs).

   This yields the DP:
   \[
   dp[n][m] = \min_{1 \le k < n}\left(2\cdot dp[k][m] + dp[n-k][m-1]\right)
   \]

3. `N ≤ 64`, `M ≤ 65`, so a cubic DP (`O(N^2 M)`) is tiny and fast.

4. To **print moves**, we must **reconstruct** the decisions:
   - For each `(n, m)` find a `k` that achieves the optimal `dp[n][m]`.
   - Actually simulate pegs as stacks so we can print `"atop <top>"` correctly.

---

## 3) Full solution approach

### Step A — Compute DP table

Let `dp[i][j]` = minimal moves (according to Frame–Stewart recurrence) to move `i` disks using `j` pegs.

Base:
- `dp[0][j] = 0` for all `j`
- `dp[1][j] = 1` for `j ≥ 2`

Transition for `i ≥ 2`, `j ≥ 3`:
- Try all `k = 1..i-1`:
  \[
  dp[i][j] = \min(dp[i][j], 2\cdot dp[k][j] + dp[i-k][j-1])
  \]

We also keep `choice[i][j] = k` that attains the minimum (any one is fine).

### Step B — Simulate pegs and print an optimal sequence

Maintain `pegs[1..M]` as stacks containing disk sizes; initially peg 1 has `N, N-1, ..., 1` (top is `1`).

Define:
- `move_disc(a, b)`: pops top from peg `a`, prints the required line (with `atop` if needed), then pushes onto peg `b`.

Recursive reconstruction:
`solve_rec(cnt, num_pegs, from, to)` moves the **top `cnt` disks** from `from` to `to` using `num_pegs` pegs:
- If `cnt == 0`: return
- If `cnt == 1`: `move_disc(from, to)`
- Else:
  - `k = choice[cnt][num_pegs]`
  - Choose an auxiliary peg `aux` not equal to `from` or `to` that can legally accept the next moved disk (in practice, an empty peg always exists for this problem setting).
  - Perform:
    1. `solve_rec(k, num_pegs, from, aux)`
    2. `solve_rec(cnt-k, num_pegs-1, from, to)`
    3. `solve_rec(k, num_pegs, aux, to)`

Finally:
- Print `dp[N][M]`
- Initialize stacks
- Call `solve_rec(N, M, 1, M)`

---

## 4) C++ implementation (with detailed comments)

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

/*
  Towers of Hanoi with M pegs (M >= 4), N disks (N <= 64).
  Use Frame–Stewart DP:
    dp[n][m] = min_{1<=k<n} (2*dp[k][m] + dp[n-k][m-1])
  Then reconstruct one optimal sequence and print moves,
  including "atop X" if destination is non-empty.
*/

static const long long INF = (1LL << 60);

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

    int N, M;
    cin >> N >> M;

    // dp[i][j] = minimal move count for i disks, j pegs
    vector<vector<long long>> dp(N + 1, vector<long long>(M + 1, INF));
    // choice[i][j] = k achieving dp[i][j]
    vector<vector<int>> choice(N + 1, vector<int>(M + 1, -1));

    // Base cases
    for (int j = 1; j <= M; j++) dp[0][j] = 0;
    for (int j = 2; j <= M; j++) dp[1][j] = 1;

    // Fill DP
    for (int i = 2; i <= N; i++) {
        for (int j = 3; j <= M; j++) {
            for (int k = 1; k < i; k++) {
                long long cand = 2 * dp[k][j] + dp[i - k][j - 1];
                if (cand < dp[i][j]) {
                    dp[i][j] = cand;
                    choice[i][j] = k;
                }
            }
        }
    }

    cout << dp[N][M] << "\n";

    // Simulate pegs as stacks (top is back / stack::top).
    vector<vector<int>> pegs(M + 1);
    pegs[1].reserve(N);
    for (int d = N; d >= 1; d--) pegs[1].push_back(d);

    auto move_disc = [&](int from, int to) {
        int disc = pegs[from].back();
        // Print required format:
        cout << "move " << disc << " from " << from << " to " << to;
        if (!pegs[to].empty()) {
            cout << " atop " << pegs[to].back();
        }
        cout << "\n";

        // Perform the move in simulation
        pegs[from].pop_back();
        pegs[to].push_back(disc);
    };

    // Recursively output moves according to choice[][].
    function<void(int,int,int,int)> rec = [&](int cnt, int num_pegs, int from, int to) {
        if (cnt == 0) return;
        if (cnt == 1) {
            move_disc(from, to);
            return;
        }

        int k = choice[cnt][num_pegs];
        // Defensive: in correct DP, k must exist for cnt>=2, num_pegs>=3.
        if (k < 1) k = 1;

        // Pick an auxiliary peg different from from/to.
        // Additionally, it must be legal to place the first moved disk there.
        // The first disk to move is the current top of 'from'.
        int topDisc = pegs[from].back();
        int aux = -1;
        for (int p = 1; p <= M; p++) {
            if (p == from || p == to) continue;
            if (pegs[p].empty() || pegs[p].back() > topDisc) {
                aux = p;
                break;
            }
        }

        // 1) Move k disks from 'from' to 'aux' using all num_pegs
        rec(k, num_pegs, from, aux);
        // 2) Move remaining cnt-k from 'from' to 'to' using num_pegs-1
        rec(cnt - k, num_pegs - 1, from, to);
        // 3) Move k from 'aux' to 'to' using all num_pegs
        rec(k, num_pegs, aux, to);
    };

    rec(N, M, 1, M);
    return 0;
}
```

---

## 5) Python implementation (with detailed comments)

```python
import sys
sys.setrecursionlimit(10**7)

"""
Frame–Stewart DP for multi-peg Hanoi:

dp[n][m] = min_{1<=k<n} (2*dp[k][m] + dp[n-k][m-1])

Then reconstruct moves using choice[n][m] and a peg simulation
to print "atop X" when needed.
"""

def solve() -> None:
    data = sys.stdin.read().strip().split()
    N = int(data[0])
    M = int(data[1])

    INF = 10**30

    dp = [[INF] * (M + 1) for _ in range(N + 1)]
    choice = [[-1] * (M + 1) for _ in range(N + 1)]

    # Base cases
    for j in range(1, M + 1):
        dp[0][j] = 0
    for j in range(2, M + 1):
        dp[1][j] = 1

    # DP fill
    for n in range(2, N + 1):
        for m in range(3, M + 1):
            best = INF
            best_k = -1
            for k in range(1, n):
                cand = 2 * dp[k][m] + dp[n - k][m - 1]
                if cand < best:
                    best = cand
                    best_k = k
            dp[n][m] = best
            choice[n][m] = best_k

    # Peg simulation: pegs[i] is a stack, top is last element
    pegs = [[] for _ in range(M + 1)]
    for d in range(N, 0, -1):
        pegs[1].append(d)

    out = []

    def move_disc(frm: int, to: int) -> None:
        disc = pegs[frm][-1]
        if pegs[to]:
            out.append(f"move {disc} from {frm} to {to} atop {pegs[to][-1]}")
        else:
            out.append(f"move {disc} from {frm} to {to}")
        pegs[frm].pop()
        pegs[to].append(disc)

    def rec(cnt: int, num_pegs: int, frm: int, to: int) -> None:
        if cnt == 0:
            return
        if cnt == 1:
            move_disc(frm, to)
            return

        k = choice[cnt][num_pegs]
        if k < 1:
            k = 1  # safety fallback (shouldn't happen with valid DP states)

        # Choose an auxiliary peg that can accept the next moved disk legally.
        top_disc = pegs[frm][-1]
        aux = None
        for p in range(1, M + 1):
            if p == frm or p == to:
                continue
            if (not pegs[p]) or pegs[p][-1] > top_disc:
                aux = p
                break
        if aux is None:
            raise RuntimeError("No legal auxiliary peg found (should not happen).")

        rec(k, num_pegs, frm, aux)
        rec(cnt - k, num_pegs - 1, frm, to)
        rec(k, num_pegs, aux, to)

    # Output
    rec(N, M, 1, M)
    sys.stdout.write(str(dp[N][M]) + "\n")
    sys.stdout.write("\n".join(out) + ("\n" if out else ""))

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

---

If you want, I can also provide a version that explicitly handles `m == 3` with the classical `2^n - 1` recurrence (not needed here because `M ≥ 4`), or a variant that stores the actual auxiliary peg choices to avoid searching for `aux` during reconstruction.