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<map>
#include<set>
#include<stack>
#include<queue>
#include<string>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define pb push_back
using namespace std;
const int P=139;
const int mod=998244353;
const int maxn=1e7+5;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int minn=0xc0c0c0c0;
ll k,m,n,x,ans,f[maxn],pre[maxn];
int main()
{
    scanf("%lld%lld%lld",&n,&x,&m);
    for(int i=0;i<m;i++)
    	f[i]=-1;
    int a=0;
    ans=0;
    while(n--)
    {
    	f[x]=a++;
    	pre[x]=ans;
    	ans+=x;
    	x=x*x%m;
    	if(f[x]>=0)
    	{
    		int b=a-f[x];
    		ans+=(ans-pre[x])*(n/b);
    		n%=b;
		}
	}
	printf("%lld\n",ans);
    return 0;
}