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 int long long
int32_t main(){
	int n,x,m;
	cin>>n>>x>>m;
	vector<int>first(m,-1);
	vector<int>pre(2*m,0);
	int ans = 0;
	for(int i=1;i<=n;i++){
		if(first[x]>=0){
			int loopsize = i-first[x];
			int loopsum = pre[i-1]-pre[first[x]-1];
			
			int loops = (n-i)/loopsize;
			
			ans += loops*loopsum;
			i+=loopsize*loops;
			for(int j=0;j<m;j++){
				first[j]=-1;
			}
		}
		if(i<pre.size()){
			pre[i]=pre[i-1]+x;
			first[x]=i;
		}
		
		ans += x;
		x = (x*x)%m;
	}
	
	cout<<ans<<endl;
}

