#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
long long dp1[505], dp[505][505], fact[505], dp2[505];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  long long p;
  cin >> p;
  dp[0][0] = 1;
  dp1[0] = 1;
  dp1[1] = 1;
  dp1[2] = 2;
  dp[1][0] = 1;
  dp[2][0] = 1;
  dp[2][1] = 2;
  fact[0] = 1;
  for (int i = 1; i <= 500; i++) fact[i] = (1LL * fact[i - 1] * i) % p;
  for (int i = 1; i <= 500; i++) {
    dp2[i] = fact[i];
    for (int j = 1; j < i; j++) {
      dp2[i] = (dp2[i] - (fact[j] * dp2[i - j]) % p + p) % p;
    }
  }
  for (int i = 3; i <= 500; i++) {
    dp[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      dp[i][j] = dp[i - 1][j];
      for (int k = 2; k <= i && j - k + 1 >= 0; k++) {
        dp[i][j] = (dp[i][j] + (dp[i - k][j - k + 1] * fact[k]) % p) % p;
      }
    }
  }
  for (int i = 3; i <= 500; i++) {
    for (int j = 1; j < i; j++) {
      long long w3 = (2LL * fact[j] * dp2[i - j]) % p;
      dp1[i] = (dp1[i] + w3) % p;
    }
    for (int j = 1; j < i && i - j > 3; j++) {
      dp1[i] = (dp1[i] + (dp[i][j] * dp1[i - j]) % p) % p;
    }
    dp1[i] = (fact[i] - dp1[i] + p) % p;
  }
  for (int i = 1; i <= t; i++) {
    int n;
    cin >> n;
    cout << dp1[n] << '\n';
  }
}

### 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;
const int N = 100005;
long long dp1[505], dp[505][505], fact[505], dp2[505];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  long long p;
  cin >> p;
  dp[0][0] = 1;
  dp1[0] = 1;
  dp1[1] = 1;
  dp1[2] = 2;
  dp[1][0] = 1;
  dp[2][0] = 1;
  dp[2][1] = 2;
  fact[0] = 1;
  for (int i = 1; i <= 500; i++) fact[i] = (1LL * fact[i - 1] * i) % p;
  for (int i = 1; i <= 500; i++) {
    dp2[i] = fact[i];
    for (int j = 1; j < i; j++) {
      dp2[i] = (dp2[i] - (fact[j] * dp2[i - j]) % p + p) % p;
    }
  }
  for (int i = 3; i <= 500; i++) {
    dp[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      dp[i][j] = dp[i - 1][j];
      for (int k = 2; k <= i && j - k + 1 >= 0; k++) {
        dp[i][j] = (dp[i][j] + (dp[i - k][j - k + 1] * fact[k]) % p) % p;
      }
    }
  }
  for (int i = 3; i <= 500; i++) {
    for (int j = 1; j < i; j++) {
      long long w3 = (2LL * fact[j] * dp2[i - j]) % p;
      dp1[i] = (dp1[i] + w3) % p;
    }
    for (int j = 1; j < i && i - j > 3; j++) {
      dp1[i] = (dp1[i] + (dp[i][j] * dp1[i - j]) % p) % p;
    }
    dp1[i] = (fact[i] - dp1[i] + p) % p;
  }
  for (int i = 1; i <= t; i++) {
    int n;
    cin >> n;
    cout << dp1[n] << '\n';
  }
}

```