#include <bits/stdc++.h>
const int MAXN = 410;
const int N = 400;
int mod = 998244353;
void reduce(int& x) { x += x >> 31 & mod; }
int mul(int a, int b) { return (long long)a * b % mod; }
void fma(int& x, int y, int z) { x = ((long long)y * z + x) % mod; }
void fms(int& x, int y, int z) { x = ((long long)(mod - y) * z + x) % mod; }
int fac[MAXN];
int f[MAXN], g[MAXN], h[MAXN];
int pr[MAXN][MAXN];
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  int T;
  std::cin >> T >> mod;
  for (int i = *fac = 1; i != MAXN; ++i) fac[i] = mul(fac[i - 1], i);
  pr[0][0] = 1;
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= i; ++j)
      for (int k = 1; k <= i; ++k) fma(pr[i][j], pr[i - k][j - 1], fac[k]);
  f[1] = 1, f[2] = 1, f[3] = 3;
  h[1] = 1, h[2] = 2;
  for (int i = 4; i <= N; ++i) {
    static int cv[MAXN];
    memset(cv, 0, N + 1 << 2);
    cv[0] = 1;
    for (int j = 1; j <= i; ++j)
      for (int k = 1; k < i && k <= j; ++k) fma(cv[j], cv[j - k], f[k] + g[k]);
    f[i] = cv[i];
    reduce(g[i] = fac[i] - mul(f[i], 2));
    h[i] = g[i];
    for (int j = 4; j < i; ++j) fms(h[i], pr[i][j], h[j]);
  }
  while (T-- > 0) {
    int n;
    std::cin >> n;
    std::cout << h[n] << std::endl;
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following 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. 

### Response

```cpp
#include <bits/stdc++.h>
const int MAXN = 410;
const int N = 400;
int mod = 998244353;
void reduce(int& x) { x += x >> 31 & mod; }
int mul(int a, int b) { return (long long)a * b % mod; }
void fma(int& x, int y, int z) { x = ((long long)y * z + x) % mod; }
void fms(int& x, int y, int z) { x = ((long long)(mod - y) * z + x) % mod; }
int fac[MAXN];
int f[MAXN], g[MAXN], h[MAXN];
int pr[MAXN][MAXN];
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  int T;
  std::cin >> T >> mod;
  for (int i = *fac = 1; i != MAXN; ++i) fac[i] = mul(fac[i - 1], i);
  pr[0][0] = 1;
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= i; ++j)
      for (int k = 1; k <= i; ++k) fma(pr[i][j], pr[i - k][j - 1], fac[k]);
  f[1] = 1, f[2] = 1, f[3] = 3;
  h[1] = 1, h[2] = 2;
  for (int i = 4; i <= N; ++i) {
    static int cv[MAXN];
    memset(cv, 0, N + 1 << 2);
    cv[0] = 1;
    for (int j = 1; j <= i; ++j)
      for (int k = 1; k < i && k <= j; ++k) fma(cv[j], cv[j - k], f[k] + g[k]);
    f[i] = cv[i];
    reduce(g[i] = fac[i] - mul(f[i], 2));
    h[i] = g[i];
    for (int j = 4; j < i; ++j) fms(h[i], pr[i][j], h[j]);
  }
  while (T-- > 0) {
    int n;
    std::cin >> n;
    std::cout << h[n] << std::endl;
  }
  return 0;
}

```