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)

//E - Sequence Sum
//Hiia
#include <stdio.h>
#include <stdlib.h>//rannsuu, zettaichi(absolute)
#include <time.h>
//srand((unsigned int) time(0));
#include <math.h>//koudona keisann
#include <string.h>//hairetsu
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
typedef long long ll;
//#define N 998244353
#define M 1000000007
#define pi 3.1415926535897932384626433832

int gcd(int m,int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
int lcm(int m,int n){if((0==m)||(0==n))return 0;return(m*n/gcd(m, n));}
int fact(int x){if(x==1) return(x);else return(x*fact(x-1));}//Kaijou x!
int ruijou(int base,int exponent){int i,answer=1;i=exponent;while(i!=0){answer=answer*base;--i;}return answer;}//ruijou base^exponent

long long int llgcd(long long int m,long long int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
long long int lllcm(long long int m,long long int n){if((0==m)||(0==n))return 0;return(m*n/llgcd(m, n));}
long long int llfact(long long int x){if(x==1) return(x);else return(x*llfact(x-1));}//Kaijou
int llruijou(long long int base,long long int exponent){long long int i,answer=1;i=exponent;while(i!=0){answer=answer*base;--i;}return answer;}//ruijou base^exponent

int main()
{
  ll n,x,m,ans,a[100010]={};
  scanf("%lld %lld %lld", &n,&x,&m);

  ans=x;
  a[1]=x;
  for(ll i = 2; i <= n; i++) {
    a[i]=(a[i-1]*a[i-1])%m;
    ans+=a[i];
  }

  printf("%lld\n", ans);

  return 0;
}