求逆元。

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 3e6+5;
int n,p,fac[N] = {1};
signed main()
{
	cin>>n>>p;
	for(int i = 1;i<=n;i++)
		cout<<(fac[i] = fac[p%i]*(p-p/i)%p)<<'\n';
	return 0;
}