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;
#define ll long long
#define ff first
#define ss second
int main(){
	ll n, x, m, ans = 0, s = 0, l = 0, rep;
	cin >> n >> x >> m;
	bool b[m];
	fill(b, b + m, false);
	ll cur = x;
	for (int i = 0; i < m; i++){
		if (b[cur]) rep = cur;
		b[cur] = true;
		(cur *= cur) %= m;
	}
	cur = x;
	while (cur != rep){
		ans += cur;
		(cur *= cur) %= m, n--;
		if (n == 0) break;
	}
	do{
		s += cur;
		(cur *= cur) %= m, l++;
	}while (cur != rep);
	for (int i = 0; i < n % l; i++){
		ans += cur;
		(cur *= cur) %= m;
	}
	ans += s * (n / l);
	printf("%lld\n", ans);
}