<|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 ≤ 10^9).

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:

```text
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;

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;
};

int n, mod;

void read() { cin >> n >> mod; }

void solve() {
    // dp [pos] [dir][value] = dp[pos - 1][!dir][x]      for x in [0, value)
    //                                                   if dir == 0
    //                                                   for x in [value, pos]
    //                                                   if dir == 1

    if(n == 1) {
        cout << 1 % mod << '\n';
        return;
    }

    vector<vector<int>> last_dp(2), dp(2);
    last_dp[0] = {1 % mod};
    last_dp[1] = {1 % mod};

    for(int pos = 1; pos < n; pos++) {
        // dir = 1
        dp[1].resize(pos + 1);
        dp[1][pos] = 0; 
        for(int value = pos - 1; value >= 0; value--) {
            dp[1][value] = last_dp[0][value] + dp[1][value + 1];
            if(dp[1][value] >= mod) {
                dp[1][value] -= mod;
            }
        }
        // dir = 0
        dp[0].resize(pos + 1);
        dp[0][0] = 0;
        for(int value = 1; value <= pos; value++) {
            dp[0][value] = last_dp[1][value - 1] + dp[0][value - 1];
            if(dp[0][value] >= mod) {
                dp[0][value] -= mod;
            }
        }
        swap(last_dp, dp);
    }

    int ans = 0;
    for(int value = 0; value <= n - 1; value++) {
        ans = (ans + last_dp[0][value]) % mod;
        ans = (ans + last_dp[1][value]) % mod;
    }

    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();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

5. Python implementation with detailed comments

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

    # If n = 1, only one permutation
    if n == 1:
        print(1 % mod)
        return

    # last_dp[0] and last_dp[1] are lists of size = current length
    # last_dp[d][v] = number of prefixes of length L ending at rank v, last step direction d
    last_dp = [ [1],  # at length=1, only rank 0, count=1, direction=down
                [1] ] # and also count=1 for direction=up (base)

    # Build up from prefix length=1 to n
    for length in range(2, n+1):
        pos = length - 1  # zero-based index for new dp size
        dp_up   = [0] * length  # dp for direction=up
        dp_down = [0] * length  # dp for direction=down

        # Compute dp_up[v] = sum(last_dp[0][x] for x in [v..pos-1]), do suffix sums
        suffix = 0
        # We treat last_dp[0] as having size pos
        for v in range(pos-1, -1, -1):
            suffix = (suffix + last_dp[0][v]) % mod
            dp_up[v] = suffix
        # dp_up[pos] stays 0 (no x ≥ pos)

        # Compute dp_down[v] = sum(last_dp[1][x] for x in [0..v-1]), do prefix sums
        prefix = 0
        for v in range(1, length):
            prefix = (prefix + last_dp[1][v-1]) % mod
            dp_down[v] = prefix
        # dp_down[0] is 0 (no x < 0)

        # Prepare for next iteration
        last_dp = [dp_down, dp_up]

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

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