#include <bits/stdc++.h>
using namespace std;
long long p;
const int maxn = 409;
long long fact[maxn];
long long alpha[maxn][maxn];
long long a[maxn];
long long b[maxn];
long long mod(long long& x) {
  x %= p;
  if (x < 0) x += p;
  return x;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int t;
  cin >> t >> p;
  fact[0] = 1;
  for (int i = 1; i < (maxn); ++i) {
    fact[i] = fact[i - 1] * i;
    mod(fact[i]);
  }
  alpha[0][0] = 1;
  for (int i = 1; i < (maxn); ++i) {
    for (int j = 1; j < (maxn); ++j) {
      if (j > i) {
        continue;
      }
      for (int k = 1; k < (i + 1); ++k) {
        alpha[i][j] += fact[k] * alpha[i - k][j - 1];
        mod(alpha[i][j]);
      }
    }
  }
  a[0] = 0;
  b[0] = 0;
  a[1] = 1;
  b[1] = 1;
  for (int n = 2; n < (maxn); ++n) {
    b[n] = fact[n];
    for (int k = 1; k < (n); ++k) {
      b[n] -= b[k] * fact[n - k];
      mod(b[n]);
    }
    a[n] = 2 * b[n] - fact[n];
    for (int k = 4; k < (n); ++k) {
      a[n] -= alpha[n][k] * a[k];
      mod(a[n]);
    }
  }
  a[2] = 2;
  for (int i = 0; i < (t); ++i) {
    int nn;
    cin >> nn;
    cout << a[nn] << endl;
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 p;
const int maxn = 409;
long long fact[maxn];
long long alpha[maxn][maxn];
long long a[maxn];
long long b[maxn];
long long mod(long long& x) {
  x %= p;
  if (x < 0) x += p;
  return x;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int t;
  cin >> t >> p;
  fact[0] = 1;
  for (int i = 1; i < (maxn); ++i) {
    fact[i] = fact[i - 1] * i;
    mod(fact[i]);
  }
  alpha[0][0] = 1;
  for (int i = 1; i < (maxn); ++i) {
    for (int j = 1; j < (maxn); ++j) {
      if (j > i) {
        continue;
      }
      for (int k = 1; k < (i + 1); ++k) {
        alpha[i][j] += fact[k] * alpha[i - k][j - 1];
        mod(alpha[i][j]);
      }
    }
  }
  a[0] = 0;
  b[0] = 0;
  a[1] = 1;
  b[1] = 1;
  for (int n = 2; n < (maxn); ++n) {
    b[n] = fact[n];
    for (int k = 1; k < (n); ++k) {
      b[n] -= b[k] * fact[n - k];
      mod(b[n]);
    }
    a[n] = 2 * b[n] - fact[n];
    for (int k = 4; k < (n); ++k) {
      a[n] -= alpha[n][k] * a[k];
      mod(a[n]);
    }
  }
  a[2] = 2;
  for (int i = 0; i < (t); ++i) {
    int nn;
    cin >> nn;
    cout << a[nn] << endl;
  }
}

```