#include <bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 405;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  string s = "/home/jslijin/jslijin/code/";
  freopen((s + "code.in").c_str(), "r", stdin);
  freopen((s + "code.out").c_str(), "w", stdout);
}
long long I[N], B[N][N], A[N], fac[N];
int n, p, T;
void init() {
  int n = 400;
  fac[0] = 1;
  for (int i = (1); i <= (n); ++i) fac[i] = fac[i - 1] * i % p;
  for (int i = (1); i <= (n); ++i) {
    I[i] = fac[i];
    for (int k = (1); k <= (i - 1); ++k)
      I[i] -= I[k] * fac[i - k] % p, I[i] %= p;
  }
  B[0][0] = 1;
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (i); ++j)
      for (int t = (j - 1); t <= (i - 1); ++t)
        B[i][j] += B[t][j - 1] * fac[i - t] % p, B[i][j] %= p;
  for (int i = (1); i <= (n); ++i) {
    A[i] = (2 * I[i] - fac[i]) % p;
    for (int k = (3); k <= (i - 1); ++k) A[i] -= B[i][k] * A[k] % p, A[i] %= p;
  }
}
int main() {
  scanf("%d%d", &T, &p);
  init();
  while (T--) scanf("%d", &n), printf("%lld\n", n == 2 ? 2 : (A[n] + p) % p);
  return 0;
}

### 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>
using namespace std;
const int oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 405;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  string s = "/home/jslijin/jslijin/code/";
  freopen((s + "code.in").c_str(), "r", stdin);
  freopen((s + "code.out").c_str(), "w", stdout);
}
long long I[N], B[N][N], A[N], fac[N];
int n, p, T;
void init() {
  int n = 400;
  fac[0] = 1;
  for (int i = (1); i <= (n); ++i) fac[i] = fac[i - 1] * i % p;
  for (int i = (1); i <= (n); ++i) {
    I[i] = fac[i];
    for (int k = (1); k <= (i - 1); ++k)
      I[i] -= I[k] * fac[i - k] % p, I[i] %= p;
  }
  B[0][0] = 1;
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (i); ++j)
      for (int t = (j - 1); t <= (i - 1); ++t)
        B[i][j] += B[t][j - 1] * fac[i - t] % p, B[i][j] %= p;
  for (int i = (1); i <= (n); ++i) {
    A[i] = (2 * I[i] - fac[i]) % p;
    for (int k = (3); k <= (i - 1); ++k) A[i] -= B[i][k] * A[k] % p, A[i] %= p;
  }
}
int main() {
  scanf("%d%d", &T, &p);
  init();
  while (T--) scanf("%d", &n), printf("%lld\n", n == 2 ? 2 : (A[n] + p) % p);
  return 0;
}

```