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

const int mx=100010;
int sp[mx][35];
ll cum[mx][35];
int main(){
	ll n,x,m;
	cin>>n>>x>>m;
	for(int i=0;i<m;i++){
		sp[i][0]=(i*1LL*i)%m;
		cum[i][0]=sp[i][0];
	}
	for(int j=1;j<35;j++){
		for(int i=0;i<m;i++){
			sp[i][j]=i;
			cum[i][j]+=cum[sp[i][j]][j-1];
			sp[i][j]=sp[sp[i][j]][j-1];

			cum[i][j]+=cum[sp[i][j]][j-1];
			sp[i][j]=sp[sp[i][j]][j-1];
		}
	}
	ll curr=x,an=x;
	for(int i=35;i>=0;i--){
		if((n-1)&(1LL<<i)){
			an+=cum[curr][i];
			curr=sp[curr][i];
		}
	}
	cout<<an<<endl;
}