#include <bits/stdc++.h>
using namespace std;
const int maxN = 405;
int dp[maxN][maxN];
int fact[maxN];
int f[maxN];
int cnk[maxN][maxN];
int tst;
int mod;
int n;
int sum(int a, int b) {
  int s = a + b;
  if (s >= mod) s -= mod;
  return s;
}
int sub(int a, int b) {
  int s = a - b;
  if (s < 0) s += mod;
  return s;
}
int mult(int a, int b) { return (1LL * a * b) % mod; }
void solve() { cin >> n; }
int main() {
  cin >> tst >> mod;
  cnk[0][0] = 1;
  fact[0] = 1;
  int lim = 400;
  for (int i = 1; i <= lim; i++) {
    fact[i] = mult(fact[i - 1], i);
    cnk[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      cnk[i][j] = sum(cnk[i - 1][j], cnk[i - 1][j - 1]);
    }
  }
  dp[0][0] = 1;
  for (int cnt = 0; cnt <= lim; cnt++) {
    for (int s = 0; s <= lim; s++) {
      for (int add = 1; add <= lim; add++) {
        if (s + add <= lim) {
          dp[cnt + 1][s + add] =
              sum(dp[cnt + 1][s + add], mult(dp[cnt][s], fact[add]));
        }
      }
    }
  }
  for (int i = 3; i <= lim; i++) {
    f[i] = fact[i];
    for (int j = 3; j < i; j++) {
      f[i] = sub(f[i], mult(f[j], dp[j][i]));
    }
    int q = 0;
    for (int k = 1; k <= i + 1; k++) {
      if (k & 1) {
        q = sum(q, dp[k + 1][i]);
      } else {
        q = sub(q, dp[k + 1][i]);
      }
    }
    q = mult(q, 2);
    f[i] = sub(f[i], q);
  }
  for (int i = 1; i <= tst; i++) {
    int n;
    cin >> n;
    if (n <= 2)
      cout << n << '\n';
    else
      cout << f[n] << '\n';
  }
  return 0;
}

### Prompt

Generate 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;
const int maxN = 405;
int dp[maxN][maxN];
int fact[maxN];
int f[maxN];
int cnk[maxN][maxN];
int tst;
int mod;
int n;
int sum(int a, int b) {
  int s = a + b;
  if (s >= mod) s -= mod;
  return s;
}
int sub(int a, int b) {
  int s = a - b;
  if (s < 0) s += mod;
  return s;
}
int mult(int a, int b) { return (1LL * a * b) % mod; }
void solve() { cin >> n; }
int main() {
  cin >> tst >> mod;
  cnk[0][0] = 1;
  fact[0] = 1;
  int lim = 400;
  for (int i = 1; i <= lim; i++) {
    fact[i] = mult(fact[i - 1], i);
    cnk[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      cnk[i][j] = sum(cnk[i - 1][j], cnk[i - 1][j - 1]);
    }
  }
  dp[0][0] = 1;
  for (int cnt = 0; cnt <= lim; cnt++) {
    for (int s = 0; s <= lim; s++) {
      for (int add = 1; add <= lim; add++) {
        if (s + add <= lim) {
          dp[cnt + 1][s + add] =
              sum(dp[cnt + 1][s + add], mult(dp[cnt][s], fact[add]));
        }
      }
    }
  }
  for (int i = 3; i <= lim; i++) {
    f[i] = fact[i];
    for (int j = 3; j < i; j++) {
      f[i] = sub(f[i], mult(f[j], dp[j][i]));
    }
    int q = 0;
    for (int k = 1; k <= i + 1; k++) {
      if (k & 1) {
        q = sum(q, dp[k + 1][i]);
      } else {
        q = sub(q, dp[k + 1][i]);
      }
    }
    q = mult(q, 2);
    f[i] = sub(f[i], q);
  }
  for (int i = 1; i <= tst; i++) {
    int n;
    cin >> n;
    if (n <= 2)
      cout << n << '\n';
    else
      cout << f[n] << '\n';
  }
  return 0;
}

```