#include <bits/stdc++.h>
using namespace std;
long long ways[401][401];
int main() {
  long long n = 400;
  long long MOD;
  int T;
  cin >> T;
  cin >> MOD;
  vector<long long> fact(n + 1);
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % MOD;
  }
  vector<long long> inc(n + 1);
  inc[1] = 0;
  for (long long len = 1; len <= n; len++) {
    vector<long long> tmp(len + 1, 0);
    tmp[0] = 1;
    for (long long f = 1; f <= len; f++) {
      for (long long g = 1; g <= f; g++) {
        if (g == len) continue;
        tmp[f] += tmp[f - g] * (fact[g] - inc[g]);
        tmp[f] %= MOD;
      }
    }
    inc[len] = tmp[len];
  }
  vector<long long> ans(n + 1);
  ans[0] = 0;
  ways[0][0] = 1;
  for (long long a = 1; a <= n; a++) {
    for (long long b = 1; b <= n; b++) {
      for (long long c = 0; c < a; c++) {
        ways[a][b] += ways[c][b - 1] * fact[a - c];
        ways[a][b] %= MOD;
      }
    }
  }
  for (long long len = 1; len <= n; len++) {
    ans[len] = fact[len];
    ans[len] -= 2 * inc[len];
    for (long long cnt = 2; cnt < len; cnt++) {
      ans[len] -= ans[cnt] * ways[len][cnt];
      ans[len] %= MOD;
    }
  }
  ans[2] = 2;
  for (int t = 0; t < T; t++) {
    int nn;
    cin >> nn;
    long long f = ans[nn];
    f %= MOD;
    if (f < 0) f += MOD;
    cout << f << '\n';
  }
}

### Prompt

Please formulate 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;
long long ways[401][401];
int main() {
  long long n = 400;
  long long MOD;
  int T;
  cin >> T;
  cin >> MOD;
  vector<long long> fact(n + 1);
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % MOD;
  }
  vector<long long> inc(n + 1);
  inc[1] = 0;
  for (long long len = 1; len <= n; len++) {
    vector<long long> tmp(len + 1, 0);
    tmp[0] = 1;
    for (long long f = 1; f <= len; f++) {
      for (long long g = 1; g <= f; g++) {
        if (g == len) continue;
        tmp[f] += tmp[f - g] * (fact[g] - inc[g]);
        tmp[f] %= MOD;
      }
    }
    inc[len] = tmp[len];
  }
  vector<long long> ans(n + 1);
  ans[0] = 0;
  ways[0][0] = 1;
  for (long long a = 1; a <= n; a++) {
    for (long long b = 1; b <= n; b++) {
      for (long long c = 0; c < a; c++) {
        ways[a][b] += ways[c][b - 1] * fact[a - c];
        ways[a][b] %= MOD;
      }
    }
  }
  for (long long len = 1; len <= n; len++) {
    ans[len] = fact[len];
    ans[len] -= 2 * inc[len];
    for (long long cnt = 2; cnt < len; cnt++) {
      ans[len] -= ans[cnt] * ways[len][cnt];
      ans[len] %= MOD;
    }
  }
  ans[2] = 2;
  for (int t = 0; t < T; t++) {
    int nn;
    cin >> nn;
    long long f = ans[nn];
    f %= MOD;
    if (f < 0) f += MOD;
    cout << f << '\n';
  }
}

```