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;
long long T, mod, n, fac[500], f[500], g[500], dp[500], h[500];
void upd(long long &x, long long y) { x = (x + y) % mod; }
int main() {
  scanf("%lld%lld", &T, &mod);
  n = 400;
  fac[0] = fac[1] = 1;
  f[1] = 1;
  for (long long i = 2; i <= n; ++i) {
    fac[i] = fac[i - 1] * i % mod;
    memset(dp, 0, sizeof dp);
    dp[0] = 1;
    for (long long j = 1; j <= i; ++j) {
      for (long long k = 1; k <= j && k < i; ++k) {
        upd(dp[j], dp[j - k] * (f[k] + g[k]));
      }
    }
    g[i] = dp[i];
    f[i] = (fac[i] - g[i] * 2 % mod + mod) % mod;
  }
  memset(dp, 0, sizeof dp);
  for (long long i = 1; i <= n; ++i) dp[i] = (f[i] + g[i] * 2) % mod;
  for (long long i = 1; i <= n; ++i) h[i] = f[i];
  for (long long i = 2; i <= n; ++i) {
    for (long long j = n; j; --j) {
      dp[j] = 0;
      for (long long k = 1; k <= j; ++k)
        upd(dp[j], dp[j - k] * (f[k] + g[k] * 2));
      if (j > i) upd(h[j], -h[i] * dp[j] % mod + mod);
    }
  }
  h[2] = 2;
  while (T--) {
    scanf("%lld", &n);
    printf("%lld\n", h[n]);
  }
  return 0;
}