<|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.

489. Extremal Permutations
Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard



A member ai of the sequence a1, a2, ·s, an is called a  if either ai > ai-1 and ai > ai+1 (local maximum) or ai < ai-1 and ai < ai+1 (local minimum). A sequence p1, p2, ·s, pn is called a  of the integers from 1 to n if each of the integers appears in the sequence exactly once. A permutation is called  if each member (except the first and the last) is a local extreme.

Compute the total number of extremal permutations of the integers from 1 to n and output the result modulo m.

Input
The first and only line of the input file contains the integers n () and m (1 ≤ m ≤ 109).

Output
The output file should contain a single integer, the remainder from division of the total number of extremal permutations of integers from 1 to n by the given integer m.

Example(s)
sample input
sample output
3 10
4

sample input
sample output
3 3
1



Note. The extremal permutations of 1·s3 are (1, 3, 2), (2, 1, 3), (2, 3, 1) and (3, 1, 2).

<|response|>
1. Abridged problem statement  
Count the number of permutations p of {1,2,…,n} such that for every i=2…n−1, pᵢ is either a strict local maximum (pᵢ>pᵢ₋₁ and pᵢ>pᵢ₊₁) or a strict local minimum (pᵢ<pᵢ₋₁ and pᵢ<pᵢ₊₁). Output this count modulo a given integer m.

2. Key observations  
- Such permutations are exactly the “alternating” permutations (also called up‐down or zig-zag permutations).  
- We can build them incrementally by inserting one new element at a time into a growing sequence.  
- It suffices to track, after k elements are placed (k from 1 to n), two things:  
  • The direction of the last step, d ∈ {0,1}, where 0 means “last step was down” (pₖ < pₖ₋₁) and 1 means “last step was up” (pₖ > pₖ₋₁).  
  • The rank r of the last element among the k placed elements (0 ≤ r ≤ k−1), i.e. how many are smaller than it in the current partial permutation.  
- When we insert the (k+1)-th new largest value, we choose one of the k+1 possible insertion positions; that determines its new rank r_new. The comparison between r_new and the old r tells us whether the new step is “up” or “down.”  
- We must alternate steps, so if the previous step was down (d=0), the new step must be up (d_new=1), and vice versa.

3. Full solution approach  
Let dp[k][d][r] = number of alternating prefixes of length k+1 (i.e. k+1 elements used) whose last step direction is d and whose last element has rank r among those k+1.  
- Base case (k=0, length=1): there is exactly one way to place the first element, and it can be thought of as having come from “up” or “down,” so  
  dp[0][0][0] = dp[0][1][0] = 1.  
- Transition: to go from length k to k+1, we insert a new largest element (so we go from k elements to k+1). We choose its insertion rank r_new in [0…k].  
  • If we want the new step to be “up” (d_new=1), we must have r_new > r_old. Since r_old runs over 0…k−1, this is equivalent to summing dp[k−1][0][r_old] over all r_old < r_new.  
  • If we want the new step to be “down” (d_new=0), we must have r_new < r_old, i.e. sum dp[k−1][1][r_old] over r_old ≥ r_new.  
Thus  
  dp[k][1][r_new] = ∑_{r_old=0…r_new−1} dp[k−1][0][r_old],  
  dp[k][0][r_new] = ∑_{r_old=r_new…k−1} dp[k−1][1][r_old].  
To compute these in O(k) time per k (and O(k) space) we maintain prefix sums (for the first) and suffix sums (for the second).  
- After we build up to k = n−1 (length n), the answer is  
  ∑_{r=0…n−1} (dp[n−1][0][r] + dp[n−1][1][r]) mod m.  
Overall time complexity is O(n²) and memory O(n).

4. C++ implementation with detailed comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

/*
  We compute dp layer by layer.
  last_dp[d][r] stores dp[k-1][d][r] at the beginning of each iteration,
  where k is the current length we are extending to.
*/
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    long long mod;
    cin >> n >> mod;

    // Edge case: for n=1 there is exactly one permutation.
    if (n == 1) {
        cout << (1 % mod) << "\n";
        return 0;
    }

    // last_dp[0] = ways ending with a "down" step
    // last_dp[1] = ways ending with an "up" step
    // Each vector has size = current length (k), storing ranks 0..k-1
    vector<vector<long long>> last_dp(2), dp(2);

    // Base: length=1 (k=0 in 0-based), only rank=0, both directions = 1
    last_dp[0] = {1 % mod};
    last_dp[1] = {1 % mod};

    // Build from length=1 up to length=n
    // We index k from 1 to n-1 (0-based k=0 was length=1)
    for (int k = 1; k < n; k++) {
        // New length is k+1, so ranks run 0..k
        dp[0].assign(k+1, 0);
        dp[1].assign(k+1, 0);

        // Compute dp[k][1][r] = sum_{x=0..r-1} last_dp[0][x] (prefix sums)
        long long prefix = 0;
        for (int r = 0; r <= k; r++) {
            if (r > 0) {
                prefix = (prefix + last_dp[0][r-1]) % mod;
            }
            dp[1][r] = prefix;
        }

        // Compute dp[k][0][r] = sum_{x=r..k-1} last_dp[1][x] (suffix sums)
        long long suffix = 0;
        for (int r = k; r >= 0; r--) {
            if (r < k) {
                suffix = (suffix + last_dp[1][r]) % mod;
            }
            dp[0][r] = suffix;
        }

        // Move dp into last_dp for next iteration
        last_dp.swap(dp);
    }

    // Sum all possibilities of final step direction and final rank
    long long answer = 0;
    for (int r = 0; r < n; r++) {
        answer = (answer + last_dp[0][r]) % mod;
        answer = (answer + last_dp[1][r]) % mod;
    }
    cout << answer << "\n";
    return 0;
}
```

5. Python implementation with detailed comments  
```python
import sys

def main():
    data = sys.stdin.read().split()
    n, mod = map(int, data)

    # If n=1 there is exactly one permutation
    if n == 1:
        print(1 % mod)
        return

    # last_dp[0][r] = count ending with a down step at rank r
    # last_dp[1][r] = count ending with an up step at rank r
    last_dp = [
        [1],  # length=1, rank=0, down=1
        [1],  # length=1, rank=0, up=1
    ]

    # Build sequences of length 2,3,...,n
    # k is the new length minus 1 (0-based)
    for k in range(1, n):
        # We will compute dp for length = k+1, ranks 0..k
        dp_down = [0] * (k+1)
        dp_up   = [0] * (k+1)

        # dp_up[r] = sum of last_dp[0][x] for x in [0..r-1]
        prefix = 0
        for r in range(0, k+1):
            if r > 0:
                prefix = (prefix + last_dp[0][r-1]) % mod
            dp_up[r] = prefix

        # dp_down[r] = sum of last_dp[1][x] for x in [r..k-1]
        suffix = 0
        for r in range(k, -1, -1):
            if r < k:
                suffix = (suffix + last_dp[1][r]) % mod
            dp_down[r] = suffix

        last_dp = [dp_down, dp_up]

    # Sum over both directions and all ranks
    result = sum(last_dp[0]) + sum(last_dp[1])
    print(result % mod)

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