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: Accepted)

#include<bits/stdc++.h>
typedef long long int ll;
using namespace std;
#define sz 200009
#define mx 10000000000008
#define mod 1000000007
ll n,t,arr[sz],brr[sz],crr[sz],x,m;
vector<ll>v;
int main()
{
    scanf("%lld %lld %lld",&n,&x,&m);
    t=x;
    arr[x]=1;
    brr[1]=x;
    for(int i=2;i<=n;i++)
    {
        ll val=(x*x)%m;
        brr[i]=brr[i-1]+val;
        if(arr[val])
        {
            ll num=i-arr[val];
            ll sum=brr[i]-brr[arr[val]];
            ll rem=n-i;
            ll jog=(rem/num)*sum;
            ll ans=brr[i]+jog;
            ll bak=rem%num;
            for(int i=1;i<=bak;i++)
            {
                val=(val*val)%m;
                ans+=val;
            }
            printf("%lld\n",ans);
            return 0;
        }
        x=val;
        arr[val]=i;
    }
    printf("%lld\n",brr[n]);
    return 0;
}