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>
const int N = 405;
int mod, binom[N][N], factor[N];
int a[N], s[N], c[N], tc[N], itc[N];
int dp[N][N];
void up(int &x, int y) { x += y - mod, x += x >> 31 & mod; }
void up(int &x, int y, int z) { x = (x + (long long)y * z) % mod; }
void down(int &x, int y) { x -= y, x += x >> 31 & mod; }
void init(int n) {
  factor[0] = 1;
  for (int i = 1; i <= n; ++i) factor[i] = (long long)factor[i - 1] * i % mod;
  for (int i = 0; i <= n; ++i)
    for (int j = binom[i][0] = 1; j <= i; ++j)
      up(binom[i][j] = binom[i - 1][j], binom[i - 1][j - 1]);
  tc[1] = itc[1] = 1, s[1] = 1;
  tc[2] = itc[2] = 1, s[2] = 2;
  dp[0][0] = 1;
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      for (int k = 1; k <= n - i; ++k)
        up(dp[i + k][j + 1], dp[i][j], factor[k]);
  for (int i = 3; i <= n; ++i) {
    for (int j = 4; j < i; ++j) up(a[i], s[j], dp[i][j]);
    up(itc[i - 1], tc[i - 1]);
    for (int j = 1; j < i; ++j) up(itc[i], itc[j], tc[i - j]);
    c[i] = itc[i], down(tc[i] = factor[i], c[i]);
    s[i] = (factor[i] + mod * 3u - a[i] - c[i] * 2) % mod;
  }
}
int main() {
  std::ios::sync_with_stdio(0), std::cin.tie(0);
  int tc, x;
  std::cin >> tc >> mod, init(400);
  while (tc--) std::cin >> x, std::cout << s[x] << '\n';
  return 0;
}