#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using dbl = long double;
using pii = pair<int, int>;
int MOD;
void sadd(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int add(int a, int b) {
  sadd(a, b);
  return a;
}
void ssub(int &a, int b) {
  a -= b;
  if (a < 0) a += MOD;
}
int sub(int a, int b) {
  ssub(a, b);
  return a;
}
int mul(int a, int b) { return (a * 1LL * b) % MOD; }
const int N = 405;
int dp[N][N];
int fact[N];
int ans[N];
int splitPref[N];
void calcDp() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) fact[i] = mul(fact[i - 1], i);
  for (int n = 1; n < N; n++) {
    for (int k = 1; k < n; k++) {
      int cur = sub(fact[k], splitPref[k]);
      cur = mul(cur, fact[n - k]);
      sadd(splitPref[n], cur);
    }
  }
  dp[0][0] = 1;
  for (int pos = 0; pos < N; pos++)
    for (int cnt = 0; cnt <= pos; cnt++) {
      int cur = dp[pos][cnt];
      for (int k = 1; k + pos < N; k++)
        sadd(dp[pos + k][cnt + 1], mul(cur, fact[k]));
    }
  for (int k = 1; k < N; k++) {
    ans[k] = fact[k];
    for (int t = 3; t < k; t++) ssub(ans[k], mul(dp[k][t], ans[t]));
    ssub(ans[k], mul(2, splitPref[k]));
    if (k < 12) (void)0;
  }
}
void solve() {
  int t;
  scanf("%d%d", &t, &MOD);
  calcDp();
  ans[2] = 2;
  for (int i = 0; i < t; i++) {
    int n;
    scanf("%d", &n);
    printf("%d\n", ans[n]);
  }
}
int main() {
  solve();
  (void)0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
using ll = long long;
using dbl = long double;
using pii = pair<int, int>;
int MOD;
void sadd(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int add(int a, int b) {
  sadd(a, b);
  return a;
}
void ssub(int &a, int b) {
  a -= b;
  if (a < 0) a += MOD;
}
int sub(int a, int b) {
  ssub(a, b);
  return a;
}
int mul(int a, int b) { return (a * 1LL * b) % MOD; }
const int N = 405;
int dp[N][N];
int fact[N];
int ans[N];
int splitPref[N];
void calcDp() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) fact[i] = mul(fact[i - 1], i);
  for (int n = 1; n < N; n++) {
    for (int k = 1; k < n; k++) {
      int cur = sub(fact[k], splitPref[k]);
      cur = mul(cur, fact[n - k]);
      sadd(splitPref[n], cur);
    }
  }
  dp[0][0] = 1;
  for (int pos = 0; pos < N; pos++)
    for (int cnt = 0; cnt <= pos; cnt++) {
      int cur = dp[pos][cnt];
      for (int k = 1; k + pos < N; k++)
        sadd(dp[pos + k][cnt + 1], mul(cur, fact[k]));
    }
  for (int k = 1; k < N; k++) {
    ans[k] = fact[k];
    for (int t = 3; t < k; t++) ssub(ans[k], mul(dp[k][t], ans[t]));
    ssub(ans[k], mul(2, splitPref[k]));
    if (k < 12) (void)0;
  }
}
void solve() {
  int t;
  scanf("%d%d", &t, &MOD);
  calcDp();
  ans[2] = 2;
  for (int i = 0; i < t; i++) {
    int n;
    scanf("%d", &n);
    printf("%d\n", ans[n]);
  }
}
int main() {
  solve();
  (void)0;
}

```