#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int N = 3000010;

//递推打表求逆元
LL inv[N];
void getInv(LL n, LL p) {
    inv[1] = 1;
    for (int i = 2; i <= n; i++)
        inv[i] = inv[p % i] * (p - p / i) % p;
}

int main() {
    LL n, p;
    cin >> n >> p;
    getInv(n, p);
    for (int i = 1; i <= n; i++) printf("%d\n", inv[i]); //可以AC
    // cout << inv[i] << "\n"; //可以AC
    // cout << inv[i] << endl; //这里使用cout就会TLE两个点!
    return 0;
}