#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& p) {
  return (out << p.first << ' ' << p.second << endl);
}
const int INF = 1000000017;
const long long LLINF = 1000000000000000017LL;
const long double inf = 1.0 / 0.0;
const long double eps = 1e-10;
const int MOD = 1e9 + 7;
const int MOD2 = 1e9 + 9;
const int N = 405;
long long first[N] = {};
long long dp[N][N] = {};
long long inv[N][N] = {};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int q, p;
  cin >> q >> p;
  first[0] = 1;
  for (int i = 1; i < N; ++i) {
    first[i] = first[i - 1] * i % p;
  }
  for (int n = 1; n < N; ++n) {
    dp[n][1] = first[n];
    for (int k = 2; k <= n; ++k) {
      for (int m = 0; m <= n - k; ++m) {
        dp[n][k] += first[m + 1] * dp[n - m - 1][k - 1];
        dp[n][k] %= p;
      }
    }
  }
  for (int n = 1; n < N; ++n) {
    for (int k = 1; k <= n; ++k) {
      long long cur = 0;
      for (int m = 1; m < n; ++m) {
        cur += dp[n][m] * inv[m][k];
        cur %= p;
      }
      inv[n][k] = ((n == k) - cur + p);
      if (inv[n][k] >= p) {
        inv[n][k] -= p;
      }
    }
  }
  while (q--) {
    int n;
    cin >> n;
    if (n == 2) {
      cout << 2 << '\n';
    } else {
      cout << ((-inv[n][1] + 2 * (n % 2 == 0 ? -1 : 1)) % p + p) % p << '\n';
    }
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

Consider a permutation p_1, p_2, ... p_n of integers from 1 to n. We call a sub-segment p_l, p_{l+1}, ..., p_{r-1}, p_{r} of the permutation an interval if it is a reordering of some set of consecutive integers. For example, the permutation (6,7,1,8,5,3,2,4) has the intervals (6,7), (5,3,2,4), (3,2), and others. 

Each permutation has some trivial intervals — the full permutation itself and every single element. We call a permutation interval-free if it does not have non-trivial intervals. In other words, interval-free permutation does not have intervals of length between 2 and n - 1 inclusive. 

Your task is to count the number of interval-free permutations of length n modulo prime number p. 

Input

In the first line of the input there are two integers t (1 ≤ t ≤ 400) and p (10^8 ≤ p ≤ 10^9) — the number of test cases to solve and the prime modulo. In each of the next t lines there is one integer n (1 ≤ n ≤ 400) — the length of the permutation. 

Output

For each of t test cases print a single integer — the number of interval-free permutations modulo p. 

Examples

Input


4 998244353
1
4
5
9


Output


1
2
6
28146


Input


1 437122297
20


Output


67777575

Note

For n = 1 the only permutation is interval-free. For n = 4 two interval-free permutations are (2,4,1,3) and (3,1,4,2). For n = 5 — (2,4,1,5,3), (2,5,3,1,4), (3,1,5,2,4), (3,5,1,4,2), (4,1,3,5,2), and (4,2,5,1,3). We will not list all 28146 for n = 9, but for example (4,7,9,5,1,8,2,6,3), (2,4,6,1,9,7,3,8,5), (3,6,9,4,1,5,8,2,7), and (8,4,9,1,3,6,2,7,5) are interval-free.

The exact value for n = 20 is 264111424634864638. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& p) {
  return (out << p.first << ' ' << p.second << endl);
}
const int INF = 1000000017;
const long long LLINF = 1000000000000000017LL;
const long double inf = 1.0 / 0.0;
const long double eps = 1e-10;
const int MOD = 1e9 + 7;
const int MOD2 = 1e9 + 9;
const int N = 405;
long long first[N] = {};
long long dp[N][N] = {};
long long inv[N][N] = {};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int q, p;
  cin >> q >> p;
  first[0] = 1;
  for (int i = 1; i < N; ++i) {
    first[i] = first[i - 1] * i % p;
  }
  for (int n = 1; n < N; ++n) {
    dp[n][1] = first[n];
    for (int k = 2; k <= n; ++k) {
      for (int m = 0; m <= n - k; ++m) {
        dp[n][k] += first[m + 1] * dp[n - m - 1][k - 1];
        dp[n][k] %= p;
      }
    }
  }
  for (int n = 1; n < N; ++n) {
    for (int k = 1; k <= n; ++k) {
      long long cur = 0;
      for (int m = 1; m < n; ++m) {
        cur += dp[n][m] * inv[m][k];
        cur %= p;
      }
      inv[n][k] = ((n == k) - cur + p);
      if (inv[n][k] >= p) {
        inv[n][k] -= p;
      }
    }
  }
  while (q--) {
    int n;
    cin >> n;
    if (n == 2) {
      cout << 2 << '\n';
    } else {
      cout << ((-inv[n][1] + 2 * (n % 2 == 0 ? -1 : 1)) % p + p) % p << '\n';
    }
  }
  return 0;
}

```