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 nth_bit(int64_t num, int n){
    return (num >> n) & 1;
}

int main(){
    int64_t N, X, M;
    cin >> N >> X >> M;
    static int64_t sum[40][100000], nxt[40][100000];
    for(int64_t i=0; i<M; i++){
        sum[0][i] = i;
        nxt[0][i] = i*i%M;
    }
    for(int k=0; k<39; k++) for(int i=0; i<M; i++){
        sum[k+1][i] = sum[k][i] + sum[k][nxt[k][i]];
        nxt[k+1][i] = nxt[k][nxt[k][i]];
    }

    int64_t ans = 0;
    for(int k=39; k>=0; k--) if(nth_bit(N, k)){
        ans += sum[k][X];
        X = nxt[k][X];
    }
    cout << ans << endl;
    return 0;
}
