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;
const int N = 1e5 + 5;
long long n, x, m;
int vis[N] = {};
long long ans[N] = {};
int main() {
    cin >> n >> x >> m;
    x %= m;
    ans[1] = x;
    vis[x] = 1;
    int r = -1, d;
    for (int i = 2; i <= n; i++) {
        x = x * x % m;
        ans[i] = ans[i - 1] + x;
        if (vis[x]) {
            r = vis[x];
            d = i - vis[x];
            break;
        }
        vis[x] = i;
    }
    if (r == -1) cout << ans[n] << '\n';
    else {
        long long f = (n - r) / d, rr = (n - r) % d;
        cout << ans[r + rr] + (ans[r + d] - ans[r]) * f << '\n';
    }
    return 0;
}