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 mod;
int fac[405];
int f0[405], f1[405];
int g[405][405], sumg[405][405];
int ans[405];
void add(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void sub(int& x, int y) {
  x -= y;
  if (x < 0) x += mod;
}
int mul(int x, int y) { return (long long)x * y % mod; }
int main() {
  int tt;
  scanf("%d %d", &tt, &mod);
  fac[0] = 1;
  for (int i = 1; i <= 400; ++i) fac[i] = mul(fac[i - 1], i);
  for (int n = 1; n <= 400; ++n) {
    f0[n] = fac[n];
    for (int i = 2; i <= n - 1; ++i) {
      sub(f0[n], mul(f0[i], mul(fac[n - i], n - i + 1)));
    }
  }
  for (int n = 1; n <= 400; ++n) {
    f1[n] = fac[n - 1];
    for (int i = 2; i <= n - 1; ++i) {
      sub(f1[n], mul(f1[i], fac[n - i]));
    }
  }
  for (int n = 1; n <= 400; ++n) {
    int tot = 0;
    for (int k = 0; k <= n - 1; ++k) {
      for (int len = 2; len <= min(n - 1, n - k); ++len) {
        if (k == 1 && len == n - 1) {
          add(g[n][k], f0[n - 1]);
          add(g[n][k], f0[n - 1]);
          sub(g[n][k], f1[n - 1]);
          sub(g[n][k], f1[n - 1]);
        } else {
          int now = fac[n - len + 1];
          if (k) sub(now, sumg[n - len + 1][k - 1]);
          add(g[n][k], mul(f0[len], now));
        }
      }
      add(tot, g[n][k]);
      sumg[n][k] = tot;
    }
  }
  for (int n = 1; n <= 400; ++n) {
    ans[n] = fac[n];
    for (int i = 0; i <= n - 2; ++i) sub(ans[n], g[n][i]);
  }
  while (tt--) {
    int n;
    scanf("%d", &n);
    printf("%d\n", ans[n]);
  }
}