## 1. Abridged Problem Statement

Count the number of "extremal" (alternating) permutations of {1,2,…,n}, i.e. permutations p where for every i=2…n−1, either pᵢ is a strict peak (pᵢ>pᵢ₋₁ and pᵢ>pᵢ₊₁) or a strict valley (pᵢ<pᵢ₋₁ and pᵢ<pᵢ₊₁). Output the total count modulo m.

## 2. Detailed Editorial

We wish to count permutations of length n in which entries alternate up and down at every interior position. A classic way is an O(n²) DP by building the permutation element by element, tracking two pieces of state:
- d ∈ {0,1}: the direction of the last step (0 = last step was "down," 1 = last step was "up").
- v ∈ [0…i]: the rank (0-based) of the most recently placed element among the i+1 elements used so far.

Define dp[i][d][v] = number of ways to form an alternating prefix of length i+1 with last‐step direction d and ending at rank v. We start at i=0 (length=1) with dp[0][0][0] = dp[0][1][0] = 1 (the first element can be considered trivially as coming from either direction).

To add the (i+2)-th element (i from 0 to n−2), we insert a new number that is largest so far but choose its rank v_new ∈ [0…i+1]. The direction of the step from the previous element to this new one must alternate:
  - If we want an "up" step (d_new=1), the new rank must be ≥ old rank.
  - If we want a "down" step (d_new=0), the new rank must be ≤ old rank.

Thus the transitions are:

```text
dp[i+1][1][v] = sum_{x=v…i} dp[i][0][x]
dp[i+1][0][v] = sum_{x=0…v−1} dp[i][1][x]
```

We implement these sums efficiently by maintaining prefix (or suffix) sums in O(1) per state, leading to an overall O(n²) time and O(n) space. Finally, the answer is:

```text
∑_{v=0…n−1} (dp[n−1][0][v] + dp[n−1][1][v]) mod m
```

## 3. C++ Solution

```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;
}
```

## 4. Python Solution

```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()
```

## 5. Compressed Editorial

- Define dp[i][d][v] = count of alternating prefixes of length i+1 ending at relative rank v, with last step direction d.
- Base: dp[0][0][0] = dp[0][1][0] = 1.
- Transition by inserting a new largest element with chosen rank v_new:
  - For an "up" step (d=1), sum dp[i−1][0][x] over x ≥ v_new.
  - For a "down" step (d=0), sum dp[i−1][1][x] over x < v_new.
- Maintain these sums by prefix/suffix accumulation in O(1) per state so total time is O(n²), space O(n).
- Final answer = Σ_v (dp[n−1][0][v] + dp[n−1][1][v]) mod m.
