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 <cstdio>
#include <cstring>
#define max_m 100000
#define max_min_n_m max_m
using namespace std;
long long sum[max_min_n_m+1];
int vis[max_m+1];
int main(){
	long long n,x;
	int m;
	scanf("%lld%lld%d",&n,&x,&m);
	sum[1]=x,memset(vis,0,sizeof(vis));
	for(int i=1; ++i<=n; sum[i]=sum[i-1]+x,vis[x]=i){
		x=x*x%m;
		if(!x){
			printf("%lld\n",sum[i-1]);
			return 0;
		}
		if(vis[x]){
			printf("%lld\n",(sum[i-1]-sum[vis[x]-1])*((n-vis[x]+1)/(i-vis[x]))+sum[vis[x]+(n-vis[x]+1)%(i-vis[x])-1]);
			return 0;
		}
	}
	printf("%lld\n",sum[n]);
	return 0;
}