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;
typedef long double ld;
#define MAX 5000000000000000000LL
#define MIN -5000000000000000000LL

ll n, x, m, xx;
map<ll, ll> chk;
ll cou;
ll hap;
ll ans;

int main(void)
{
	scanf("%lld %lld %lld", &n, &x, &m);
	
	xx = x;
	
	chk[x] = 1;
	
	while(1)
	{
		x = (x * x) % m;
		
		if(chk[x])
			break;
		
		chk[x] = 1;
	}
	
	while(n > 0 && xx != x)
	{
		ans += xx;
		xx = (xx * xx) % m;
		n--;
	}
	
	xx = x;
	
	hap += x;
	cou++;
	x = (x * x) % m;
	
	while(x != xx)
	{
		hap += x;
		cou++;
		x = (x * x) % m;
	}
	
	ans += (n / cou) * hap;
	
	n %= cou;
	
	for(ll i = 0 ; i < n ; i++)
	{
		ans += x;
		x = (x * x) % m;
	}
	
	printf("%lld", ans);
	return 0;
}