#include <bits/stdc++.h>
using namespace std;
bool home = 1;
signed realMain();
signed main() {
  home = 0;
  if (home) {
    freopen("input", "r", stdin);
  } else {
    ios::sync_with_stdio(0);
    cin.tie(0);
  }
  realMain();
}
const int N = 400 + 7;
int q, mod, fact[N], no_pre[N], ways[N][N], ret[N];
int add(int a, int b) {
  a += b;
  if (a >= mod) return a - mod;
  if (a < 0) return a + mod;
  return a;
}
int mul(int a, int b) { return a * (long long)b % mod; }
signed realMain() {
  cin >> q >> mod;
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
  for (int i = 1; i < N; i++) {
    no_pre[i] = fact[i];
    for (int j = 1; j < i; j++) {
      no_pre[i] = add(no_pre[i], -mul(fact[j], no_pre[i - j]));
    }
  }
  ways[0][0] = 1;
  for (int n = 1; n < N; n++) {
    for (int k = 1; k < N; k++) {
      for (int last = 1; last <= n; last++) {
        ways[n][k] = add(ways[n][k], mul(fact[last], ways[n - last][k - 1]));
      }
    }
  }
  for (int i = 1; i < N; i++) {
    ret[i] = fact[i];
    for (int j = 1; j < i; j++) {
      ret[i] = add(ret[i], -mul(2, mul(fact[j], no_pre[i - j])));
    }
    for (int k = 3; k < i; k++) {
      ret[i] = add(ret[i], -mul(ways[i][k], ret[k]));
    }
  }
  ret[2] = 2;
  while (q--) {
    int n;
    cin >> n;
    cout << ret[n] << "\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;
bool home = 1;
signed realMain();
signed main() {
  home = 0;
  if (home) {
    freopen("input", "r", stdin);
  } else {
    ios::sync_with_stdio(0);
    cin.tie(0);
  }
  realMain();
}
const int N = 400 + 7;
int q, mod, fact[N], no_pre[N], ways[N][N], ret[N];
int add(int a, int b) {
  a += b;
  if (a >= mod) return a - mod;
  if (a < 0) return a + mod;
  return a;
}
int mul(int a, int b) { return a * (long long)b % mod; }
signed realMain() {
  cin >> q >> mod;
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
  for (int i = 1; i < N; i++) {
    no_pre[i] = fact[i];
    for (int j = 1; j < i; j++) {
      no_pre[i] = add(no_pre[i], -mul(fact[j], no_pre[i - j]));
    }
  }
  ways[0][0] = 1;
  for (int n = 1; n < N; n++) {
    for (int k = 1; k < N; k++) {
      for (int last = 1; last <= n; last++) {
        ways[n][k] = add(ways[n][k], mul(fact[last], ways[n - last][k - 1]));
      }
    }
  }
  for (int i = 1; i < N; i++) {
    ret[i] = fact[i];
    for (int j = 1; j < i; j++) {
      ret[i] = add(ret[i], -mul(2, mul(fact[j], no_pre[i - j])));
    }
    for (int k = 3; k < i; k++) {
      ret[i] = add(ret[i], -mul(ways[i][k], ret[k]));
    }
  }
  ret[2] = 2;
  while (q--) {
    int n;
    cin >> n;
    cout << ret[n] << "\n";
  }
  return 0;
}

```