Let us denote by f(x, m) the remainder of the Euclidean division of x by m.

Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.

Constraints

* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N X M


Output

Print \displaystyle{\sum_{i=1}^N A_i}.

Examples

Input

6 2 1001


Output

1369


Input

1000 2 16


Output

6


Input

10000000000 10 99959


Output

492443256176507
#include<bits/stdc++.h>
using namespace std;
using ll = long long;

ll n, x, m, ans, id[100100], sum[1001001];

ll f(ll n, ll x){
	ll re = 0;
	for(ll i=1;i<=n;i++){
		re += x;
		x = (x * x) % m;
	}
	return re;
}

int main(){
	cin >> n >> x >> m;

	for(ll i=1;i<=n;i++){
		if(id[x]){
			ll left = n - i + 1;
			ll sz = i - id[x];

			ll s = 0;
			for(ll j=id[x];j<i;j++) s += sum[j];

			ans += (left / sz) * s;
			ans += f(left % sz, x);
			break;
		}

		ans += x;
		id[x] = i;
		sum[i] = x;
		x = x * x % m;
	}

	cout << ans;
}