#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using D = double;
using uint = unsigned int;
template <typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll MOD;
ll add(ll x, ll y) {
  x += y;
  if (x >= MOD) return x - MOD;
  return x;
}
ll sub(ll x, ll y) {
  x -= y;
  if (x < 0) return x + MOD;
  return x;
}
ll mult(ll x, ll y) { return (x * y) % MOD; }
const int N = 404;
ll f[N];
ll A[N];
ll P[N];
ll B[N];
ll dp[N][N];
int main() {
  int t;
  scanf("%d%lld", &t, &MOD);
  f[0] = 1;
  for (int i = 1; i < N; i++) f[i] = mult(f[i - 1], i);
  for (int i = 0; i < N; i++) {
    B[i] = f[i];
    for (int j = 1; j < i; j++) B[i] = sub(B[i], mult(B[j], f[i - j]));
  }
  for (int i = 0; i < N; i++) {
    P[i] = f[i];
    for (int j = 2; j < i; j++) P[i] = sub(P[i], mult(P[j], f[i - j + 1]));
  }
  dp[1][1] = 1;
  for (int n = 0; n < N; n++) {
    A[n] = P[n];
    if (n >= 3) A[n] = sub(A[n], mult(2, B[n - 1]));
    for (int m = 1; m < n; m++)
      for (int k = 0; k < n; k++)
        dp[n][k + 1] = add(dp[n][k + 1], mult(dp[n - m][k], f[m]));
    for (int k = 3; k < n; k++) A[n] = sub(A[n], mult(dp[n][k], A[k]));
  }
  while (t--) {
    int x;
    scanf("%d", &x);
    printf("%lld\n", A[x]);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
using namespace std;
using ll = long long;
using ld = long double;
using D = double;
using uint = unsigned int;
template <typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll MOD;
ll add(ll x, ll y) {
  x += y;
  if (x >= MOD) return x - MOD;
  return x;
}
ll sub(ll x, ll y) {
  x -= y;
  if (x < 0) return x + MOD;
  return x;
}
ll mult(ll x, ll y) { return (x * y) % MOD; }
const int N = 404;
ll f[N];
ll A[N];
ll P[N];
ll B[N];
ll dp[N][N];
int main() {
  int t;
  scanf("%d%lld", &t, &MOD);
  f[0] = 1;
  for (int i = 1; i < N; i++) f[i] = mult(f[i - 1], i);
  for (int i = 0; i < N; i++) {
    B[i] = f[i];
    for (int j = 1; j < i; j++) B[i] = sub(B[i], mult(B[j], f[i - j]));
  }
  for (int i = 0; i < N; i++) {
    P[i] = f[i];
    for (int j = 2; j < i; j++) P[i] = sub(P[i], mult(P[j], f[i - j + 1]));
  }
  dp[1][1] = 1;
  for (int n = 0; n < N; n++) {
    A[n] = P[n];
    if (n >= 3) A[n] = sub(A[n], mult(2, B[n - 1]));
    for (int m = 1; m < n; m++)
      for (int k = 0; k < n; k++)
        dp[n][k + 1] = add(dp[n][k + 1], mult(dp[n - m][k], f[m]));
    for (int k = 3; k < n; k++) A[n] = sub(A[n], mult(dp[n][k], A[k]));
  }
  while (t--) {
    int x;
    scanf("%d", &x);
    printf("%lld\n", A[x]);
  }
  return 0;
}

```