#include <bits/stdc++.h>
using namespace std;
int get() {
  char ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-') {
    int s = 0;
    while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
    return -s;
  }
  int s = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
  return s;
}
const int N = 405;
const int L = 400;
int mo;
long long f[N], g[N][N], s[N];
long long js[N], inv[N];
long long h[N];
int main() {
  int T = get();
  mo = get();
  js[0] = js[1] = 1;
  for (int i = 2; i <= L; i++) js[i] = js[i - 1] * i % mo;
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= L; i++) inv[i] = 1ll * (mo - mo / i) * inv[mo % i] % mo;
  for (int i = 2; i <= L; i++) inv[i] = inv[i] * inv[i - 1] % mo;
  g[0][0] = 1;
  for (int i = 0; i <= L - 1; i++)
    for (int x = 0; x <= L; x++)
      if (g[i][x])
        for (int y = 1; y <= L - x; y++)
          g[i + 1][x + y] = (g[i + 1][x + y] + g[i][x] * js[y]) % mo;
  h[2] = 1;
  for (int i = 1; i <= L; i++) {
    f[i] = js[i];
    for (int j = 2; j <= i - 1; j++)
      f[i] = (f[i] + mo - f[j] * (g[j - 1][i - 1]) % mo) % mo;
    s[i] = f[i];
    for (int j = 3; j <= i - 1; j++)
      s[i] = (s[i] + mo - s[j] * f[i - j + 1] % mo) % mo;
    if (i > 2) {
      h[i] = (f[i - 1] + mo - h[i - 1]) % mo;
      s[i] = (s[i] + mo - h[i] * 2 % mo) % mo;
    }
  }
  for (; T; T--) {
    int n = get();
    cout << s[n] << endl;
  }
  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;
int get() {
  char ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-') {
    int s = 0;
    while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
    return -s;
  }
  int s = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
  return s;
}
const int N = 405;
const int L = 400;
int mo;
long long f[N], g[N][N], s[N];
long long js[N], inv[N];
long long h[N];
int main() {
  int T = get();
  mo = get();
  js[0] = js[1] = 1;
  for (int i = 2; i <= L; i++) js[i] = js[i - 1] * i % mo;
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= L; i++) inv[i] = 1ll * (mo - mo / i) * inv[mo % i] % mo;
  for (int i = 2; i <= L; i++) inv[i] = inv[i] * inv[i - 1] % mo;
  g[0][0] = 1;
  for (int i = 0; i <= L - 1; i++)
    for (int x = 0; x <= L; x++)
      if (g[i][x])
        for (int y = 1; y <= L - x; y++)
          g[i + 1][x + y] = (g[i + 1][x + y] + g[i][x] * js[y]) % mo;
  h[2] = 1;
  for (int i = 1; i <= L; i++) {
    f[i] = js[i];
    for (int j = 2; j <= i - 1; j++)
      f[i] = (f[i] + mo - f[j] * (g[j - 1][i - 1]) % mo) % mo;
    s[i] = f[i];
    for (int j = 3; j <= i - 1; j++)
      s[i] = (s[i] + mo - s[j] * f[i - j + 1] % mo) % mo;
    if (i > 2) {
      h[i] = (f[i - 1] + mo - h[i - 1]) % mo;
      s[i] = (s[i] + mo - h[i] * 2 % mo) % mo;
    }
  }
  for (; T; T--) {
    int n = get();
    cout << s[n] << endl;
  }
  return 0;
}

```