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>
int t, p;
int cn = 2, tn;
long long c[500];
const int spec[] = {1, 2, 0};
int main() {
  scanf("%d%d", &t, &p);
  c[1] = p - 1;
  for (; t--;) {
    scanf("%d", &tn);
    if (tn <= 3) {
      printf("%d\n", spec[tn - 1]);
      continue;
    }
    if (tn < cn) {
      printf("%lld\n", (c[tn] + ((tn & 1) ? (2) : (-2))) % p);
      continue;
    }
    for (int n = cn; n <= tn; ++n) {
      for (int i = 1; i < n - 1; ++i)
        c[n] =
            (c[n] +
             c[i + 1] * (((i + 1) * c[n - i] + (i + 2) * c[n - i - 1]) % p)) %
            p;
      c[n] = (c[n] - c[n - 1] - c[n - 1] + p + p) % p;
    }
    printf("%lld\n", (c[tn] + ((tn & 1) ? (2) : (-2))) % p);
    cn = tn + 1;
  }
}