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 long long N = 405;
long long T, M, fac[N], dp1[N], dp2[N][N], Ans[N], n;
signed main() {
  scanf("%lld%lld", &T, &M);
  fac[0] = 1;
  for (long long i = 1; i < N; i++) fac[i] = fac[i - 1] * i % M;
  for (long long i = 1; i < N; i++) {
    dp1[i] = fac[i];
    for (long long j = 1; j < i; j++) (dp1[i] -= fac[i - j] * dp1[j]) %= M;
  }
  dp2[0][0] = 1;
  for (long long i = 1; i < N; i++)
    for (long long j = 1; j < N; j++)
      for (long long k = 1; k <= j; k++)
        (dp2[i][j] += dp2[i - 1][j - k] * fac[k]) %= M;
  for (long long i = 1; i < N; i++) {
    Ans[i] = fac[i];
    for (long long j = 1; j < i; j++) (Ans[i] -= 2 * fac[i - j] * dp1[j]) %= M;
    for (long long j = 3; j < i; j++) (Ans[i] -= dp2[j][i] * Ans[j]) %= M;
  }
  Ans[2] = 2 % M;
  while (T--) {
    scanf("%lld", &n);
    printf("%lld\n", (Ans[n] + M) % M);
  }
}