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<stdio.h>
int main(void)
{
    long long n,x,m,a,b,c,i=0,f=0,t,j;
    int box[100001]={0};
    scanf("%lld %lld %lld",&n,&x,&m);
    for(i=0;i<n;)
    {
        i++;
 //       printf("%d %lld %lld\n",box[x],x,f);
        if (box[x]==0) box[x]=i;
        else {
            b=n/i;
            c=n%i;
            f+=f*(b-1);
            for(j=0;j<b;j++)
            {
                f+=x;
                x=(x*x)%m;
            }
            printf("%lld",f);
            return 0;
        }
        f+=x;
        x=(x*x)%m;
        if(x==0) break;
    }
    printf("%lld",f);
    return 0;
}