#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    ll M;
    cin >> T >> M;
    vector<int> Ns(T);
    int maxN = 0;
    for(int i = 0; i < T; i++){
        cin >> Ns[i];
        maxN = max(maxN, Ns[i]);
    }

    // 1) factorial mod M
    vector<ll> fact(maxN+1);
    fact[0] = 1;
    for(int i = 1; i <= maxN; i++){
        fact[i] = fact[i-1] * i % M;
    }

    // 2) involutions a[n]: a[0]=1, a[1]=1, a[n]=a[n-1]+(n-1)*a[n-2]
    vector<ll> a(maxN+1);
    a[0] = 1;
    if(maxN >= 1) a[1] = 1;
    for(int n = 2; n <= maxN; n++){
        a[n] = (a[n-1] + (ll)(n-1) * a[n-2]) % M;
    }

    // 3) pow2 up to floor(maxN/2) for c[n]
    int halfN = maxN/2;
    vector<ll> pow2(halfN+1);
    pow2[0] = 1;
    for(int i = 1; i <= halfN; i++){
        pow2[i] = (pow2[i-1] * 2) % M;
    }

    // 4) P[m] = product_{k=1..m}(4k-2) mod M for d[n]
    int quarterN = maxN/4;
    vector<ll> P(quarterN+1);
    P[0] = 1;
    for(int m = 1; m <= quarterN; m++){
        // 4*m - 2 fits in ll
        P[m] = (P[m-1] * (4LL*m - 2)) % M;
    }

    // 5) H[m] = number of involutions that are also 180°-symmetric on floor(n/2) pairs
    //    H[0]=1, H[1]=2, H[m]=2*H[m-1] + 2*(m-1)*H[m-2]
    vector<ll> H(halfN+1);
    H[0] = 1;
    if(halfN >= 1) H[1] = 2 % M;
    for(int m = 2; m <= halfN; m++){
        H[m] = (2 * H[m-1] + 2LL*(m-1) * H[m-2]) % M;
    }

    // 6) 答えを計算
    //   c[n] = fact[floor(n/2)] * pow2[floor(n/2)]
    //   d[n] = (n%4==0 || n%4==1) ? P[floor(n/4)] : 0
    //   h[n] = H[floor(n/2)]
    //   S[n] = 8*fact[n] - 8*a[n] - 4*c[n] - 2*d[n] + 6*h[n]  (mod M)
    //
    for(int n : Ns){
        ll fn = fact[n];
        ll an = a[n];
        int m2 = n/2;
        ll cn = (fact[m2] * pow2[m2]) % M;
        int m4 = n/4;
        ll dn = (n%4==0 || n%4==1) ? P[m4] : 0;
        ll hn = H[m2];

        ll S = 0;
        S = (S + 8LL * fn) % M;
        S = (S - 8LL * an % M + M) % M;
        S = (S - 4LL * cn % M + M) % M;
        S = (S - 2LL * dn % M + M) % M;
        S = (S + 6LL * hn) % M;
        cout << S << "\n";
    }
    return 0;
}