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 main()
{
    long long N, X, M;
    cin >> N >> X >> M;

    vector<pair<long long, long long>> first(M, make_pair(0, -1));

    long long current = 0;
    for (long long i = 1; i <= N; i++)
    {
        X %= M;
        current += X;
        if (first[X].second == -1)
        {
            first[X] = make_pair(current, i);
        }
        else
        {
            long long diff = i - first[X].second;
            long long loop = max(0ll, (N - i) / diff - 1);
            current += loop * (current - first[X].first);
            i += loop * diff;
        }
        X *= X;
    }
    cout << current << endl;
}
