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: Wrong Answer)

#include<bits/stdc++.h>

const int N=106;
int a[N];
int b[N];
int n;

bool judge(int x){
    for(int i=x;i<=x+2;++i)
    if(a[i]!=b[i])return false;
    return true;
}
int main(){
    std::cin>>n;
    for(int i=1;i<=n;++i){
        std::cin>>a[i]>>b[i];
    }
    bool flag=false;
    if(n>=3){
        for(int i=1;i<=n-2;++i)
        if(judge(i)){
            flag=true;
            break;
        }
    }
    std::cout<<(flag?"Yes":"No")<<'\n';
    return 0;
}