#include <bits/stdc++.h>
using namespace std;
const int MAXN = 510;
int MOD = 1000000007;
void add(int &a, int b) {
  if ((a += b) >= MOD) a -= MOD;
}
void sub(int &a, int b) {
  if ((a -= b) < 0) a += MOD;
}
int mul(int a, int b) { return (int)((a * 1LL * b) % MOD); }
int mpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
int minv(int x) { return mpow(x, MOD - 2); }
int f[MAXN];
int a[MAXN];
int b[MAXN][MAXN];
int c[MAXN];
int main() {
  int t;
  scanf("%d %d", &t, &MOD);
  f[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    f[i] = mul(f[i - 1], i);
  }
  for (int i = 1; i < MAXN; i++) {
    c[i] = f[i];
    for (int j = 1; j < i; j++) {
      sub(c[i], mul(c[j], f[i - j]));
    }
  }
  b[0][0] = 1;
  for (int i = 1; i < MAXN; i++) {
    for (int j = 1; j < MAXN; j++) {
      for (int k = 1; k <= j; k++) {
        add(b[i][j], mul(f[k], b[i - 1][j - k]));
      }
    }
  }
  a[1] = 1;
  a[2] = 2;
  for (int i = 3; i < MAXN; i++) {
    a[i] = f[i];
    for (int j = 1; j < i; j++) {
      sub(a[i], mul(c[j], f[i - j]));
      sub(a[i], mul(c[j], f[i - j]));
    }
    for (int k = 3; k < i; k++) {
      sub(a[i], mul(a[k], b[k][i]));
    }
  }
  while (t-- > 0) {
    int n;
    scanf("%d", &n);
    printf("%d\n", a[n]);
  }
}

### 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;
const int MAXN = 510;
int MOD = 1000000007;
void add(int &a, int b) {
  if ((a += b) >= MOD) a -= MOD;
}
void sub(int &a, int b) {
  if ((a -= b) < 0) a += MOD;
}
int mul(int a, int b) { return (int)((a * 1LL * b) % MOD); }
int mpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
int minv(int x) { return mpow(x, MOD - 2); }
int f[MAXN];
int a[MAXN];
int b[MAXN][MAXN];
int c[MAXN];
int main() {
  int t;
  scanf("%d %d", &t, &MOD);
  f[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    f[i] = mul(f[i - 1], i);
  }
  for (int i = 1; i < MAXN; i++) {
    c[i] = f[i];
    for (int j = 1; j < i; j++) {
      sub(c[i], mul(c[j], f[i - j]));
    }
  }
  b[0][0] = 1;
  for (int i = 1; i < MAXN; i++) {
    for (int j = 1; j < MAXN; j++) {
      for (int k = 1; k <= j; k++) {
        add(b[i][j], mul(f[k], b[i - 1][j - k]));
      }
    }
  }
  a[1] = 1;
  a[2] = 2;
  for (int i = 3; i < MAXN; i++) {
    a[i] = f[i];
    for (int j = 1; j < i; j++) {
      sub(a[i], mul(c[j], f[i - j]));
      sub(a[i], mul(c[j], f[i - j]));
    }
    for (int k = 3; k < i; k++) {
      sub(a[i], mul(a[k], b[k][i]));
    }
  }
  while (t-- > 0) {
    int n;
    scanf("%d", &n);
    printf("%d\n", a[n]);
  }
}

```