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 int N = 405;
int T, p, fac[N], g[N], f[N][N], ans[N];
inline int read() {
  char c = getchar();
  int ans = 0;
  while (c < 48 || c > 57) c = getchar();
  while (c >= 48 && c <= 57)
    ans = (ans << 3) + (ans << 1) + (c ^ 48), c = getchar();
  return ans;
}
inline void write(int x) {
  int num = 0;
  char sc[15];
  if (!x) sc[num = 1] = 48;
  while (x) sc[++num] = x % 10 + 48, x /= 10;
  while (num) putchar(sc[num--]);
  putchar('\n');
}
int main() {
  T = read(), p = read();
  fac[0] = fac[1] = g[1] = f[1][1] = ans[1] = 1;
  for (register int i = 2; i <= 400; ++i) {
    g[i] = f[i][1] = ans[i] = fac[i] = 1ll * fac[i - 1] * i % p;
    for (register int j = 1; j < i; ++j)
      g[i] = (g[i] - 1ll * g[j] * fac[i - j] % p + p) % p,
      ans[i] = (ans[i] - 2ll * g[j] * fac[i - j] % p + p) % p;
    for (register int j = 2; j <= i; ++j)
      for (register int k = j - 1; k < i; ++k)
        f[i][j] = (f[i][j] + 1ll * f[k][j - 1] * fac[i - k]) % p;
    for (register int j = 3; j < i; ++j)
      ans[i] = (ans[i] - 1ll * f[i][j] * ans[j] % p + p) % p;
  }
  ans[2] = 2;
  while (T--) write(ans[read()]);
  return 0;
}