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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vb = vector<bool>;
using vs = vector<string>;
int MOD;
struct modnum {
  int v;
  modnum() : v(0) {}
  modnum(ll _v) : v(_v % MOD) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend ostream& operator<<(ostream& o, const modnum& n) { return o << n.v; }
  friend istream& operator<<(istream& i, modnum& n) {
    ll v;
    i >> v;
    n = modnum(v);
    return i;
  }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum& operator+=(const modnum& o) {
    v += o.v;
    if (v >= MOD) v -= MOD;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    if (v < 0) v += MOD;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum operator-() {
    modnum res;
    if (v) res.v = MOD - v;
    return res;
  }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  modnum pow(int e) {
    if (e == 0) return 1;
    if (e & 1) return *this * this->pow(e - 1);
    return (*this * *this).pow(e / 2);
  }
};
using mn = modnum;
using vmn = vector<mn>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int T;
  cin >> T >> MOD;
  const int MAXV = 401;
  vmn fact(MAXV);
  fact[0] = 1;
  for (int i = 1; i < MAXV; i++) fact[i] = fact[i - 1] * i;
  vmn nopref(MAXV);
  nopref[1] = 1;
  for (int i = 2; i < MAXV; i++) {
    nopref[i] = fact[i];
    for (int j = 1; j < i; j++) nopref[i] -= nopref[j] * fact[i - j];
  }
  vector<vmn> sect(MAXV, vmn(MAXV, 0));
  sect[0][0] = 1;
  sect[1][1] = 1;
  for (int i = 2; i < MAXV; i++) {
    for (int s = 1; s <= i; s++) {
      for (int l = 1; l <= i; l++) sect[i][s] += sect[i - l][s - 1] * fact[l];
    }
  }
  vmn good(MAXV);
  good[1] = 1;
  for (int i = 2; i < MAXV; i++) {
    good[i] = fact[i];
    for (int ps = 1; ps < i; ps++) {
      if (ps > 1 || i - ps > 1) good[i] -= 2 * nopref[ps] * fact[i - ps];
    }
    for (int ss = 3; ss < i; ss++) {
      good[i] -= sect[i][ss] * good[ss];
    }
  }
  for (int t = 0; t < T; t++) {
    int n;
    cin >> n;
    cout << good[n] << "\n";
  }
  return 0;
}