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. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 505;
int f[N], g[N], fac[N], n, mo;
int F[N][N], G[N][N];
void init() {
  n = 400;
  g[1] = 1;
  f[1] = 0;
  fac[0] = 1;
  for (int i = (int)(1); i <= (int)(n); i++) fac[i] = 1ll * fac[i - 1] * i % mo;
  F[0][0] = F[1][1] = 1;
  G[0][0] = G[1][1] = 1;
  for (int i = (int)(2); i <= (int)(n); i++) {
    for (int j = (int)(1); j <= (int)(i - 1); j++)
      for (int k = (int)(1); k <= (int)(j); k++)
        F[i][k + 1] =
            (F[i][k + 1] + 1ll * F[j][k] * (fac[i - j] + mo - f[i - j])) % mo;
    for (int j = (int)(2); j <= (int)(i); j++) f[i] = (f[i] + F[i][j]) % mo;
    for (int j = (int)(0); j <= (int)(i - 1); j++)
      for (int k = (int)(0); k <= (int)(j); k++)
        G[i][k + 1] = (G[i][k + 1] + 1ll * G[j][k] * fac[i - j]) % mo;
    g[i] = (fac[i] + mo - 2ll * f[i] % mo) % mo;
    for (int j = (int)(2); j <= (int)(i - 1); j++)
      g[i] = (g[i] + mo - 1ll * g[j] * G[i][j] % mo) % mo;
    F[i][1] = (F[i][1] + fac[i] + 1ll * mo - f[i]) % mo;
  }
}
int main() {
  int T;
  scanf("%d%d", &T, &mo);
  init();
  while (T--) {
    int x;
    scanf("%d", &x);
    printf("%d\n", x == 2 ? 2 : g[x]);
  }
}