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 f[401], ans[401], t, P, _mul[401], invmul[401], g[401][401];
inline int mul(const int &a, const int &b) { return 1ll * a * b % P; }
inline int add(int a, const int &b) {
  a += b;
  return (a >= P) ? a - P : a;
}
inline int sub(int a, const int &b) {
  a -= b;
  return (a < 0) ? a + P : a;
}
int qsm(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
int C(int n, int m) { return mul(mul(_mul[n], invmul[m]), invmul[n - m]); }
int main() {
  scanf("%d%d", &t, &P);
  ans[1] = 1, ans[2] = 2;
  f[1] = 1, f[2] = 1;
  _mul[0] = 1, _mul[1] = 1, _mul[2] = 2;
  invmul[0] = 1, invmul[1] = 1, invmul[2] = (P + 1) >> 1;
  g[1][1] = 1, g[2][1] = 2, g[2][2] = 1;
  for (int i = 3; i <= 400; i++) {
    _mul[i] = mul(_mul[i - 1], i);
    invmul[i] = qsm(_mul[i], P - 2);
    for (int j = 1; j < i; j++)
      f[i] = add(f[i], mul((j == 1 ? 1 : sub(_mul[j], f[j])), _mul[i - j]));
    for (int j = 2; j <= i; j++) {
      for (int k = 1; k < i; k++)
        g[i][j] = add(g[i][j], mul(g[k][j - 1], _mul[i - k]));
    }
    g[i][1] = _mul[i];
    ans[i] = sub(_mul[i], mul(f[i], 2));
    for (int j = 4; j < i; j++) {
      ans[i] = sub(ans[i], mul(ans[j], g[i][j]));
    }
  }
  for (int i = 1, n; i <= t; i++) {
    scanf("%d", &n);
    printf("%d\n", ans[n]);
  }
}