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>

#include <cstdio>

#define N 100005

using namespace std;



typedef long long ll;

ll n, x, m, sum, ans, c[N], s[N];

int main()
{

    ll i;

    cin >> n >> x >> m;

    for (i = 1; i <= n; i++, x = x * x % m) {

        if (!c[x]) c[x] = i, s[x] = sum;

        else break;

        sum += x;

    }

    if (i > n) {cout << sum << endl; return 0;}

    ll t = (n - c[x] + 1) / (i - c[x]);

    ans = s[x] + (sum - s[x]) * t;

    n = (n - c[x] + 1) % (i - c[x]);

    for (i = 1; i <= n; i++, x = x * x % m) {

        ans += x;

    }

    cout << ans << endl;

    return 0;

}
