#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 - 48 + ch;
  return x;
}
const int N = 400;
int T, mo, f[N + 5][N + 5];
int fac[N + 5], g[N + 5], h[N + 5];
int main() {
  scanf("%d%d", &T, &mo);
  fac[0] = f[0][0] = 1;
  for (int i = (int)(1); i <= (int)(N); i++)
    fac[i] = g[i] = h[i] = 1ll * fac[i - 1] * i % mo;
  for (int i = (int)(1); i <= (int)(N); i++)
    for (int j = (int)(1); j <= (int)(i - 1); j++)
      g[i] = (g[i] + mo - 1ll * g[j] * fac[i - j] % mo) % mo;
  for (int i = (int)(1); i <= (int)(N); i++)
    for (int j = (int)(1); j <= (int)(i); j++)
      for (int k = (int)(1); k <= (int)(i - j + 1); k++)
        f[i][j] = (f[i][j] + 1ll * f[i - k][j - 1] * fac[k]) % mo;
  for (int i = (int)(1); i <= (int)(N); i++)
    for (int j = (int)(1); j <= (int)(i - 1); j++)
      h[i] = (h[i] + mo - 2ll * g[j] * fac[i - j] % mo) % mo;
  for (int i = (int)(1); i <= (int)(N); i++)
    for (int j = (int)(3); j <= (int)(i - 1); j++)
      h[i] = (h[i] + mo - 1ll * h[j] * f[i][j] % mo) % mo;
  h[2] = 2;
  while (T--) printf("%d\n", h[read()]);
}

### 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;
inline int read() {
  int x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 - 48 + ch;
  return x;
}
const int N = 400;
int T, mo, f[N + 5][N + 5];
int fac[N + 5], g[N + 5], h[N + 5];
int main() {
  scanf("%d%d", &T, &mo);
  fac[0] = f[0][0] = 1;
  for (int i = (int)(1); i <= (int)(N); i++)
    fac[i] = g[i] = h[i] = 1ll * fac[i - 1] * i % mo;
  for (int i = (int)(1); i <= (int)(N); i++)
    for (int j = (int)(1); j <= (int)(i - 1); j++)
      g[i] = (g[i] + mo - 1ll * g[j] * fac[i - j] % mo) % mo;
  for (int i = (int)(1); i <= (int)(N); i++)
    for (int j = (int)(1); j <= (int)(i); j++)
      for (int k = (int)(1); k <= (int)(i - j + 1); k++)
        f[i][j] = (f[i][j] + 1ll * f[i - k][j - 1] * fac[k]) % mo;
  for (int i = (int)(1); i <= (int)(N); i++)
    for (int j = (int)(1); j <= (int)(i - 1); j++)
      h[i] = (h[i] + mo - 2ll * g[j] * fac[i - j] % mo) % mo;
  for (int i = (int)(1); i <= (int)(N); i++)
    for (int j = (int)(3); j <= (int)(i - 1); j++)
      h[i] = (h[i] + mo - 1ll * h[j] * f[i][j] % mo) % mo;
  h[2] = 2;
  while (T--) printf("%d\n", h[read()]);
}

```