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>

int main()
{
    long long n;
    int x, m;
    std::cin >> n >> x >> m;

    long long ans = 0;
    bool *flg;
    flg = new bool[m]();
    for (; n; --n)
    {
        if (flg[x])
            break;
        ans += x;
        flg[x] = true;
        x = 1ll * x * x % m;
    }
    int tmp = x, cnt = 0;
    long long sum = 0;
    n--;
    x = 1ll * x * x % m;
    for (; n; --n)
    {
        cnt++;
        sum += x;
        if (x == tmp)
            break;
        x = 1ll * x * x % m;
    }
    ans += (n / cnt + 1) * sum;
    n %= cnt;
    for (; n; --n)
    {
        ans += x;
        x = 1ll * x * x % m;
    }
    std::cout << ans << std::endl;
    return 0;
}