#include <bits/stdc++.h>
using namespace std;
int T, MOD;
long long qp(long long a, long long b) {
  long long x = 1;
  a %= MOD;
  while (b) {
    if (b & 1) x = x * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return x;
}
long long fac[666666], rfac[666666], ans[666666], c[666666];
void operator*=(vector<int>& a, vector<int>& b) {
  for (int i = 0; i < 403; ++i) c[i] = 0;
  for (int i = 0; i < 403; ++i) {
    for (int j = 0; j <= i; ++j) {
      c[i] += a[j] * (long long)b[i - j];
      if (!(j & 7)) c[i] %= MOD;
    }
    c[i] %= MOD;
  }
  for (int i = 0; i < 403; ++i) a[i] = c[i];
}
int main() {
  cin >> T >> MOD;
  fac[0] = 1;
  for (int i = 1; i < 666666; ++i) fac[i] = fac[i - 1] * i % MOD;
  rfac[666666 - 1] = qp(fac[666666 - 1], MOD - 2);
  for (int i = 666666 - 1; i >= 1; --i) rfac[i - 1] = rfac[i] * i % MOD;
  vector<int> v, x, a;
  a.resize(403);
  a[0] = 1;
  x.resize(403);
  for (int i = 0; i < 403; ++i) v.push_back(fac[i + 1]);
  for (int i = 0; i < 403; ++i) {
    long long t = (i == 0);
    for (int j = 0; j < i; ++j) t = (t - v[i - j] * (long long)x[j]) % MOD;
    x[i] = t;
  }
  for (int i = 1; i <= 400; ++i) {
    a *= x;
    ans[i] = -a[i - 1] * (long long)rfac[i] % MOD * fac[i - 1] % MOD;
    if ((i + 1) & 1)
      ans[i] -= 2;
    else
      ans[i] += 2;
    if (i == 2) ans[i] = 2;
  }
  while (T--) {
    int n;
    cin >> n;
    n = ans[n];
    n = (n % MOD + MOD) % MOD;
    cout << n << "\n";
  }
}

### Prompt

Please create a solution in Cpp 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;
int T, MOD;
long long qp(long long a, long long b) {
  long long x = 1;
  a %= MOD;
  while (b) {
    if (b & 1) x = x * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return x;
}
long long fac[666666], rfac[666666], ans[666666], c[666666];
void operator*=(vector<int>& a, vector<int>& b) {
  for (int i = 0; i < 403; ++i) c[i] = 0;
  for (int i = 0; i < 403; ++i) {
    for (int j = 0; j <= i; ++j) {
      c[i] += a[j] * (long long)b[i - j];
      if (!(j & 7)) c[i] %= MOD;
    }
    c[i] %= MOD;
  }
  for (int i = 0; i < 403; ++i) a[i] = c[i];
}
int main() {
  cin >> T >> MOD;
  fac[0] = 1;
  for (int i = 1; i < 666666; ++i) fac[i] = fac[i - 1] * i % MOD;
  rfac[666666 - 1] = qp(fac[666666 - 1], MOD - 2);
  for (int i = 666666 - 1; i >= 1; --i) rfac[i - 1] = rfac[i] * i % MOD;
  vector<int> v, x, a;
  a.resize(403);
  a[0] = 1;
  x.resize(403);
  for (int i = 0; i < 403; ++i) v.push_back(fac[i + 1]);
  for (int i = 0; i < 403; ++i) {
    long long t = (i == 0);
    for (int j = 0; j < i; ++j) t = (t - v[i - j] * (long long)x[j]) % MOD;
    x[i] = t;
  }
  for (int i = 1; i <= 400; ++i) {
    a *= x;
    ans[i] = -a[i - 1] * (long long)rfac[i] % MOD * fac[i - 1] % MOD;
    if ((i + 1) & 1)
      ans[i] -= 2;
    else
      ans[i] += 2;
    if (i == 2) ans[i] = 2;
  }
  while (T--) {
    int n;
    cin >> n;
    n = ans[n];
    n = (n % MOD + MOD) % MOD;
    cout << n << "\n";
  }
}

```