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 Kotlin (1.3.71) (judge verdict: Accepted)

import java.io.*
import java.util.*
import kotlin.math.*

fun main(){
	val f = BufferedReader(InputStreamReader(System.`in`))

	var (n,x,m) = f.readLine().split(" ").map{it.toLong()}

	val values = mutableListOf<Long>()
	values.add(x)

	val hmap = HashMap<Long,Int>()
	hmap.put(x,0)
	var start = -1

	while(true){
		val add = (values[values.size-1]*values[values.size-1])%m
		//println(add)
		if(hmap.containsKey(add)){
			start = hmap.get(add)!!
			break
		}
		hmap.put(add,values.size)
		values.add(add)
	}

	if(n <= values.size){
		var answer = 0L
		for(k in 0 until n.toInt()){
			answer += values[k]
		}
		println(answer)
	} else {

		var answer = 0L
		for(k in 0 until start){
			answer += values[k]
		}

		//println(answer)

		var sum = 0L

		for(k in start until values.size){
			sum += values[k]
		}

		//println(sum)
		n-=start
		answer += sum*(n/(values.size-start))

		var remain : Long = 0L
		for(k in 0 until ((n%(values.size-start)))){
			remain += values[start+k.toInt()]
		}

		//println(remain)

		answer += remain
		println(answer)

	}
}