#include <bits/stdc++.h>
#pragma GCC optimize "-O3"
using namespace std;
const int N = 401;
int t, mod;
int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
int mul(int a, int b) { return (long long)a * b % mod; }
int qpow(int a, int exp) {
  int ans = 1;
  while (exp) {
    if (exp & 1) ans = mul(ans, a);
    a = mul(a, a);
    exp >>= 1;
  }
  return ans;
}
void Mul(int a[], int b[]) {
  for (int i = N - 1; i >= 0; --i) {
    int cur = 0;
    for (int j = 0; j <= i; ++j) cur = add(cur, mul(a[j], b[i - j]));
    a[i] = cur;
  }
}
int a[N], b[N], c[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t >> mod;
  a[0] = 1;
  for (int i = 1; i < N; ++i) a[i] = mul(a[i - 1], i + 1);
  for (int i = 0; i < N; ++i) {
    int x = (i == 0);
    for (int j = 0; j < i; ++j) x = sub(x, mul(a[i - j], b[j]));
    b[i] = x;
  }
  for (int i = 1; i < N; ++i) a[i] = 0;
  for (int i = 1; i < N; ++i) {
    Mul(a, b);
    c[i] = mul(mod - a[i - 1], qpow(i, mod - 2));
    if (i & 1)
      c[i] = add(c[i], 2);
    else
      c[i] = sub(c[i], 2);
  }
  c[2] = 2;
  while (t--) {
    int n;
    cin >> n;
    cout << c[n] << "\n";
  }
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize "-O3"
using namespace std;
const int N = 401;
int t, mod;
int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
int mul(int a, int b) { return (long long)a * b % mod; }
int qpow(int a, int exp) {
  int ans = 1;
  while (exp) {
    if (exp & 1) ans = mul(ans, a);
    a = mul(a, a);
    exp >>= 1;
  }
  return ans;
}
void Mul(int a[], int b[]) {
  for (int i = N - 1; i >= 0; --i) {
    int cur = 0;
    for (int j = 0; j <= i; ++j) cur = add(cur, mul(a[j], b[i - j]));
    a[i] = cur;
  }
}
int a[N], b[N], c[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t >> mod;
  a[0] = 1;
  for (int i = 1; i < N; ++i) a[i] = mul(a[i - 1], i + 1);
  for (int i = 0; i < N; ++i) {
    int x = (i == 0);
    for (int j = 0; j < i; ++j) x = sub(x, mul(a[i - j], b[j]));
    b[i] = x;
  }
  for (int i = 1; i < N; ++i) a[i] = 0;
  for (int i = 1; i < N; ++i) {
    Mul(a, b);
    c[i] = mul(mod - a[i - 1], qpow(i, mod - 2));
    if (i & 1)
      c[i] = add(c[i], 2);
    else
      c[i] = sub(c[i], 2);
  }
  c[2] = 2;
  while (t--) {
    int n;
    cin >> n;
    cout << c[n] << "\n";
  }
}

```