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;

const int maxn=1e5+5;
const int maxm=40;
#define ll long long

int f[maxn][maxm];
ll sum[maxn][maxm];

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);

    ll n,x,m;
    cin>>n>>x>>m;
    for(ll i=0;i<m;i++)
    {
        f[i][0]=(i*i)%m;
        sum[i][0]=f[i][0];
    }

    for(int j=1;j<maxm;j++) for(int i=0;i<m;i++)
    {
        f[i][j]=f[f[i][j-1]][j-1];
        sum[i][j]=sum[i][j-1]+sum[f[i][j-1]][j-1];
    }

    ll ans=x;n--;
    for(int i=39;i>=0;i--) if((n>>i)&1)
    {

        ans+=sum[x][i];
        x=f[x][i];
    }
    cout<<ans;
    return 0;
}