#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007LL;
long long large = 2000000000000000000LL;
int main() {
  int t;
  cin >> t >> mod;
  vector<long long> per(401, 1);
  for (long long i = 1; i <= 400; i++) per[i] = i * per[i - 1] % mod;
  vector<long long> I(401, 0);
  I[1] = 1;
  for (long long i = 2; i <= 400; i++) {
    I[i] = per[i];
    for (long long j = 1; j < i; j++) I[i] += mod - I[j] * per[i - j] % mod;
    I[i] %= mod;
  }
  vector<vector<long long> > B(401, vector<long long>(401, 0));
  B[0][0] = 1;
  for (long long i = 1; i <= 400; i++) {
    for (long long j = 1; j <= i; j++) {
      for (long long k = 1; k <= i && i - k >= j - 1; k++) {
        B[i][j] += B[i - k][j - 1] * per[k] % mod;
      }
      B[i][j] %= mod;
    }
  }
  vector<long long> A(401, 0);
  for (long long i = 1; i <= 400; i++) {
    A[i] = per[i];
    for (long long j = 1; j < i; j++) {
      A[i] += mod - I[j] * 2 % mod * per[i - j] % mod;
    }
    A[i] %= mod;
    for (long long j = 3; j < i; j++) {
      A[i] += mod - B[i][j] * A[j] % mod;
    }
    A[i] %= mod;
    if (i == 2) A[i] = 2;
  }
  while (t--) {
    int n;
    cin >> n;
    cout << A[n] << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 mod = 1000000007LL;
long long large = 2000000000000000000LL;
int main() {
  int t;
  cin >> t >> mod;
  vector<long long> per(401, 1);
  for (long long i = 1; i <= 400; i++) per[i] = i * per[i - 1] % mod;
  vector<long long> I(401, 0);
  I[1] = 1;
  for (long long i = 2; i <= 400; i++) {
    I[i] = per[i];
    for (long long j = 1; j < i; j++) I[i] += mod - I[j] * per[i - j] % mod;
    I[i] %= mod;
  }
  vector<vector<long long> > B(401, vector<long long>(401, 0));
  B[0][0] = 1;
  for (long long i = 1; i <= 400; i++) {
    for (long long j = 1; j <= i; j++) {
      for (long long k = 1; k <= i && i - k >= j - 1; k++) {
        B[i][j] += B[i - k][j - 1] * per[k] % mod;
      }
      B[i][j] %= mod;
    }
  }
  vector<long long> A(401, 0);
  for (long long i = 1; i <= 400; i++) {
    A[i] = per[i];
    for (long long j = 1; j < i; j++) {
      A[i] += mod - I[j] * 2 % mod * per[i - j] % mod;
    }
    A[i] %= mod;
    for (long long j = 3; j < i; j++) {
      A[i] += mod - B[i][j] * A[j] % mod;
    }
    A[i] %= mod;
    if (i == 2) A[i] = 2;
  }
  while (t--) {
    int n;
    cin >> n;
    cout << A[n] << endl;
  }
  return 0;
}

```