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;
#define ll long long 

ll lev[100010], vis[100010], dis[100010], st=-1, en=-1, n, m, x;

void dfs(ll src, ll d)
{
	lev[d]=src;
	vis[src]=1;
	dis[src]=d;
	ll a=(src*src)%m;
	if(vis[a]==1)
	{
		st=dis[a];
		en=d;
		return;
	}
	dfs(a, d+1);
}

int main()
{
	ll i, j;

	scanf("%lld %lld %lld", &n, &x, &m);

	dfs(x, 1);

	if(n<=en)
	{
		ll res=0;
		for(i=1; i<=n; i++) res+=lev[i];
		printf("%lld\n", res);
		return 0;
	}

	ll res=0;

	for(i=1; i<st; i++) res+=lev[i];

	n-=(st-1);

	ll unit=en-st+1, per=0;
	
	for(i=st; i<=en; i++) per+=lev[i];

	res+=per*(n/unit);
	
	n=n%unit;

	for(i=st; i<st+n; i++) res+=lev[i];

	printf("%lld\n", res);




	return 0;
}