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>
#define int long long
using namespace std;

int32_t main(){
	int n, x, m;
	cin >> n >> x >> m;
	unordered_map<int, int> mp, sum;
	int a = 0, b = 0, c = 0;
	int i = 0;
	while(mp.count(x) == 0){
		mp[x] = i;
		b += x;
		sum[i] = b;
		i++;
		x = (x * x) % m;
	}
	a = sum[mp[x]-1];
	b -= a;
	int numCycle = (n - mp[x]) / (i - mp[x]);
	b *= numCycle;
	int rem = (n - mp[x]) % (i - mp[x]);
	c = sum[mp[x] + rem - 1] - sum[mp[x] - 1];
	cout << a + b + c;
    return 0;
}