Problem:
Score : 500 points

Problem Statement

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}.

Sample Input 1

6 2 1001

Sample Output 1

1369

The sequence A begins 2,4,16,256,471,620,\ldots Therefore, the answer is 2+4+16+256+471+620=1369.

Sample Input 2

1000 2 16

Sample Output 2

6

The sequence A begins 2,4,0,0,\ldots Therefore, the answer is 6.

Sample Input 3

10000000000 10 99959

Sample Output 3

492443256176507

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
#define int long long
using namespace std;
int n,mod,x,ans;
pair <int,int> a[100005];
int a2[100005];
int fpow(int x,int y)
{
	if(y==0) return 1;
	int t=fpow(x,y/2);
	if(y%2==0) return ((t%mod)*(t%mod))%mod;
	else return ((t%mod)*(t%mod)*(x%mod))%mod;
}
signed main()
{
	cin>>n>>x>>mod;
	int pos=0,len=0,pos2=n,sum=0;
	for(int i=1;i<=n;i++)
	{
		ans+=x;
		a[x].first=1;
		a[x].second=i;
		a2[i]=x;
		x=fpow(x,2);
		if(a[x].first==1)
		{
			pos=a[x].second;
			pos2-=i;
			len=i-pos+1;
			for(int k=pos;k<=i;k++) sum+=a2[k];
			break;
		}
	}
	ans+=(pos2/len)*sum;
	pos2%=len;
	for(int i=1;i<=pos2;i++)
	{
		ans+=x;
		x=fpow(x,2);
	}
	cout<<ans;
	return 0;
}