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>
#define FOR(i,a,b) for(int i=(a),i##end=(b);i<=i##end;i++)
#define temT template<typename T>
using namespace std;
typedef long long ll;
const int N=int(1e5)+10;

int fp[N],x,m,i,l,r; ll n,res,cur;

ll SB(){
	FOR(i,0,N-10) if(fp[i]==1) x=i;
	FOR(i,1,(int)n) res+=x,x=1ll*x*x%m;
	return res;
}

int main(){
	cin>>n>>x>>m;
	for(i=1;!fp[x];x=1ll*x*x%m,i++) fp[x]=i;
	if(fp[x]>n) return cout<<SB(),0;
	FOR(j,1,N-10) if(fp[j] && fp[j]<fp[x]) res+=j;
	l=i-fp[x]; r=(n-fp[x]+1)%l;
	FOR(j,1,N-10) if(fp[j]>=fp[x]){
		if(fp[j]-fp[x]<r) res+=j;
		cur+=j;
	}
	res+=cur*((n-fp[x]+1)/l);
	return cout<<res,0;
}