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 PI 3.141592653589793238462
using namespace std;
typedef long long ll;
typedef long double db;
ll a[100005],sum[100005];
int main(){
    ll n,x,m;cin>>n>>x>>m;
    ll now=x,num=1;
    while(!a[now]){
        a[now]=num;sum[num]=sum[num-1]+now;
        num++;now=now*now%m;
    }
    ll l=a[now],r=num-1;
    if(n<=r){
        cout<<sum[n]<<endl;return 0;
    }
    n-=r;ll ans=sum[r],v=n%(r-l+1);
    cout<<ans+(n/(r-l+1))*(sum[r]-sum[l-1])+(sum[l+v-1]-sum[l-1])<<endl;
}