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;

long long n, x, m;

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin >> n >> x >> m;
	map<long long, long long> pre;
	vector<long long> arr = {0, x};
	x = (x * x) % m;
	while(pre.find(x) == pre.end()){
		pre[x] = (int)arr.size();
		arr.push_back(x);
		x = (x * x) % m;
	}
	int cyc = pre[x], len = (int)arr.size();
	for(int i = 1; i < len; i++)
		arr[i] += arr[i - 1];
	if(n < cyc - 1) cout << arr[n] << '\n';
	else{
		long long ans = arr[cyc - 1];
		n -= cyc - 1;
		ans += (arr[len - 1] - arr[cyc - 1]) * (n / (len - cyc));
		n %= len - cyc;
		ans += arr[cyc + n - 1] - arr[cyc - 1];
		cout << ans << '\n';
	}
	return 0;
}