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. 
#include <bits/stdc++.h>
using namespace std;
int T, mod;
inline int add(int a, int b) {
  if ((a += b) >= mod) a -= mod;
  return a;
}
inline int dec(int a, int b) {
  if ((a -= b) < 0) a += mod;
  return a;
}
inline int mult(int a, int b) {
  long long t = 1ll * a * b;
  if (t >= mod) t %= mod;
  return t;
}
int f[410], fac[410], I[410], B[410][410];
int main() {
  scanf("%d%d", &T, &mod);
  fac[0] = 1;
  for (int i = 1; i <= 400; i++) fac[i] = mult(fac[i - 1], i);
  for (int i = 1; i <= 400; i++) {
    I[i] = fac[i];
    for (int j = 1; j < i; j++) I[i] = dec(I[i], mult(I[j], fac[i - j]));
  }
  B[0][0] = 1;
  for (int K = 1; K <= 400; K++) {
    for (int i = K; i <= 400; i++) {
      for (int j = 1; j <= i; j++) {
        B[i][K] = add(B[i][K], mult(B[i - j][K - 1], fac[j]));
      }
    }
  }
  for (int i = 1; i <= 400; i++) {
    f[i] = fac[i];
    for (int j = 1; j < i; j++)
      f[i] = dec(f[i], mult(2, mult(I[j], fac[i - j])));
    for (int j = 3; j < i; j++) f[i] = dec(f[i], mult(B[i][j], f[j]));
  }
  f[2] = 2;
  while (T--) {
    int x;
    scanf("%d", &x);
    printf("%d\n", f[x]);
  }
  return 0;
}
