#include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
inline unsigned int R() {
  static unsigned int seed = 416;
  return seed ^= seed >> 5, seed ^= seed << 17, seed ^= seed >> 13;
}
const int N = 404;
int T, mo, n = 400, f[N], g[N][N], fac[N], ways[N];
int main() {
  read(T);
  read(mo);
  fac[0] = 1;
  for (register int i = (1); i <= (n); i++) fac[i] = 1LL * fac[i - 1] * i % mo;
  g[0][0] = 1;
  for (register int j = (1); j <= (n); j++)
    for (register int i = (j); i <= (n); i++)
      for (register int k = (1); k <= (i); k++)
        g[j][i] = (g[j][i] + 1LL * g[j - 1][i - k] * fac[k]) % mo;
  for (register int i = (1); i <= (n); i++) {
    ways[i] = fac[i];
    for (register int j = (1); j <= (i - 1); j++)
      ways[i] = (ways[i] - 1LL * ways[j] * fac[i - j] + 1LL * mo * mo) % mo;
  }
  f[1] = 1;
  f[2] = 2;
  for (register int i = (3); i <= (n); i++) {
    f[i] = fac[i];
    for (register int j = (3); j <= (i - 1); j++)
      f[i] = (f[i] - 1LL * g[j][i] * f[j] + 1LL * mo * mo) % mo;
    for (register int j = (1); j <= (i - 1); j++)
      f[i] = (f[i] - 2LL * ways[j] * fac[i - j] + 2LL * mo * mo) % mo;
  }
  while (T--) {
    int x;
    read(x);
    printf("%d\n", f[x]);
  }
  return 0;
}

### Prompt

Please formulate 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 void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
inline unsigned int R() {
  static unsigned int seed = 416;
  return seed ^= seed >> 5, seed ^= seed << 17, seed ^= seed >> 13;
}
const int N = 404;
int T, mo, n = 400, f[N], g[N][N], fac[N], ways[N];
int main() {
  read(T);
  read(mo);
  fac[0] = 1;
  for (register int i = (1); i <= (n); i++) fac[i] = 1LL * fac[i - 1] * i % mo;
  g[0][0] = 1;
  for (register int j = (1); j <= (n); j++)
    for (register int i = (j); i <= (n); i++)
      for (register int k = (1); k <= (i); k++)
        g[j][i] = (g[j][i] + 1LL * g[j - 1][i - k] * fac[k]) % mo;
  for (register int i = (1); i <= (n); i++) {
    ways[i] = fac[i];
    for (register int j = (1); j <= (i - 1); j++)
      ways[i] = (ways[i] - 1LL * ways[j] * fac[i - j] + 1LL * mo * mo) % mo;
  }
  f[1] = 1;
  f[2] = 2;
  for (register int i = (3); i <= (n); i++) {
    f[i] = fac[i];
    for (register int j = (3); j <= (i - 1); j++)
      f[i] = (f[i] - 1LL * g[j][i] * f[j] + 1LL * mo * mo) % mo;
    for (register int j = (1); j <= (i - 1); j++)
      f[i] = (f[i] - 2LL * ways[j] * fac[i - j] + 2LL * mo * mo) % mo;
  }
  while (T--) {
    int x;
    read(x);
    printf("%d\n", f[x]);
  }
  return 0;
}

```