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;
typedef long long ll;
int main(){
	ll n,x,m;
	scanf("%lld %lld %lld",&n,&x,&m);
	ll sc=0;
	ll c=  0;
	vector<ll> f(m);
	ll k = x;
	ll ini;
	ll cont = 0;
	while(1){
		if(f[k])
			break;
		f[k] = 1;
		k = k*k%m;
	}
	ini = k;
	k = x;
	ll ans = 0;
	while(k!=ini && n--){
		ans += k;
		k=k*k%m;
		cont++;
	}
	if(n == 0){
		printf("%lld\n",ans );
		return 0;
	}
	k = ini;
	cont = 0;
	sc = 0;
	do{
		cont++;
		sc+=k;
		k=k*k%m;
	}while(k!=ini);
	ans += (n/cont)*sc;
	n = n%cont;
	while(n--){
		ans += k;
		k = k*k%m;
	}
	printf("%lld\n", ans);
	return 0;
}