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;
int P, F[505], G[505], C[505][505], fac[505], ans[505];
void up(int &x) {
  if (x >= P) x -= P;
}
void down(int &x) {
  if (x < 0) x += P;
}
int main() {
  int T = 0;
  P = 1e9 + 7;
  scanf("%d%d", &T, &P);
  fac[0] = 1;
  for (int n = 1; n <= 400; n++) fac[n] = 1ll * fac[n - 1] * n % P;
  G[1] = 1;
  for (int n = 2; n <= 400; n++) {
    G[n] = fac[n];
    for (int i = 1; i < n; i++) down(G[n] -= 1ll * G[i] * fac[n - i] % P);
  }
  C[0][0] = 1;
  for (int n = 1; n <= 400; n++)
    for (int m = 1; m <= n; m++)
      for (int i = 1; i + (m - 1) <= n; i++)
        up(C[n][m] += 1ll * C[n - i][m - 1] * fac[i] % P);
  F[1] = 1;
  for (int n = 2; n <= 400; n++) {
    F[n] = fac[n];
    for (int i = 1; i < n; i++) down(F[n] -= 2ll * G[i] * fac[n - i] % P);
    for (int i = 3; i < n; i++) down(F[n] -= 1ll * C[n][i] * F[i] % P);
    if (n == 2) F[n] = 2;
  }
  while (T--) {
    int n;
    scanf("%d", &n);
    printf("%d\n", F[n]);
  }
}