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;
int main(){
	long long n,x,m;
	cin>>n>>x>>m;
	bool visited[100005]={false};
	deque<long long>deq;
	long long ret = 0;
	while(!visited[x]){
		visited[x] = true;
		deq.push_back(x);
		long long curr = x;
		curr*=curr;
		curr%=m;
		x=curr;
	}
	while(deq.front()!=x && n){
		ret+=deq.front();
		deq.pop_front();
		n--;
	}
	long long len = deq.size(); 
	long long val = 0;
	for(int i=0; i<len; i++){
		val+=deq.front();
		deq.push_back(deq.front());
		deq.pop_front();
	}
	ret+=(n/len)*val;
	n%=len;
	for(int i=0; i<n; i++){
		ret+=deq.front();
		deq.pop_front();
	}
	cout<<ret;
}