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 int long long
using namespace std;
signed main(){
    int n,x,m;
    cin>>n>>x>>m;
    map<int,int>mp,rev;
    int pend=0,tmp=x;
    for(;;pend++,(tmp*=tmp)%=m){
        if(mp.find(tmp)!=mp.end())
            break;
        mp[tmp]=pend;
        rev[pend]=tmp;
    }
    int pbeg=mp[tmp];
    int per=pend-pbeg;
    int psum=0;
    for(int i=0;i<per;i++)
        psum+=rev[i+pbeg];
    int sum=0;
    for(int i=0;i<min(n,pbeg);i++)
        sum+=rev[i];
    if(n>=pbeg){
        n-=pbeg;
        sum+=n/per*psum;
        for(int i=0;i<n%per;i++)
            sum+=rev[i+pbeg];
    }
    cout<<sum<<endl;
}
