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;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 100005;

LL a[N];
int b[N];
LL pre[N];
int main() {
    LL n, m, x;
    cin >> n >> x >> m;
    LL ans = pre[1] = a[1] = x;
    b[x] = 1;
    for (int i = 2; i <= n; i++) {
        a[i] = a[i - 1] * a[i - 1] % m;
        ans = pre[i] = pre[i - 1] + a[i];
        if (b[a[i]] > 0) {
            LL t1 = (n - b[a[i]]) / (i - b[a[i]]), t2 = (n - b[a[i]]) % (i - b[a[i]]);
            ans = pre[b[a[i]] + t2] + (pre[i] - pre[b[a[i]]]) * t1;
            break;
        }
        b[a[i]] = i;
    }
    cout << ans << endl;
    return 0;
}
