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 <iostream>
using namespace std;
typedef long long ll;

int main()
{
    ll n, x, m;
    cin >> n >> x >> m;
    int d[100005];
    for(int i = 0; i < m; i++) d[i] = -1;
    ll k;
    ll ans = 0;
    int i;
    for(i = 0; i < n; i++){
        if(d[x] >= 0){
            k = i - d[x];
            break;
        }
        d[x] = i;
        ans += x;
        x = x * x % m;
    }
    if(i == n){
        cout << ans << endl;
        return 0;
    }
    n -= i;
    ll s = 0;
    for(i = 0; i < k; i++){
        s += x;
        x = x * x % m;
    }
    ans += n / k * s;
    n %= k;
    for(int i = 0; i < n; i++){
        ans += x;
        x = x * x % m;
    }
    cout << ans << endl;
}
