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. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int myrand() { return abs((int)mt()); }
const int maxn = 400 + 5;
int p;
int fac[maxn];
int f[maxn];
int f2[maxn];
int f3[maxn];
int g[maxn][maxn];
int calcf(int n);
int calcf2(int n);
int calcg(int n, int k);
int calcf(int n) {
  int& res = f[n];
  if (~res) return res;
  res = fac[n];
  for (int i = (0); i < (n - 1); ++i) submod(res, calcg(n, i), p);
  return res;
}
int calcf2(int n) {
  int& res = f2[n];
  if (~res) return res;
  res = fac[n];
  for (int i = (2); i < (n); ++i) {
    submod(res, mult(calcf2(i), mult(fac[n - i], n - i + 1, p), p), p);
  }
  return res;
}
int calcf3(int n) {
  int& res = f3[n];
  if (~res) return res;
  res = fac[n - 1];
  for (int i = (2); i < (n); ++i) {
    submod(res, mult(calcf3(i), fac[n - i], p), p);
  }
  return res;
}
int sumg[maxn][maxn];
int calcg(int n, int k) {
  int& res = g[n][k];
  if (~res) return res;
  res = 0;
  for (int len = (2); len < (min(n, n - k + 1)); ++len) {
    if (k == 1 && len == n - 1) {
      for (int it = (0); it < (2); ++it) {
        addmod(res, calcf2(n - 1), p);
        submod(res, calcf3(n - 1), p);
      }
    } else {
      int c = calcf2(len), tot = fac[n - len + 1];
      if (k) {
        submod(tot, sumg[n - len + 1][k - 1], p);
      }
      addmod(res, mult(c, tot, p), p);
    }
  }
  return res;
}
void chemthan() {
  int test;
  cin >> test >> p;
  fac[0] = 1;
  for (int i = (1); i < (maxn); ++i) fac[i] = mult(fac[i - 1], i, p);
  memset(f, -1, sizeof(f)), memset(f2, -1, sizeof(f2)),
      memset(f3, -1, sizeof(f3)), memset(g, -1, sizeof(g));
  for (int n = (1); n < (maxn); ++n) {
    int tot = 0;
    for (int k = (0); k < (n); ++k) {
      addmod(tot, calcg(n, k), p);
      sumg[n][k] = tot;
    }
  }
  while (test--) {
    int n;
    cin >> n;
    cout << calcf(n) << "\n";
  }
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}