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: Runtime Error)

import java.util.*

fun main() {
    val pw = java.io.PrintWriter(System.out)
    Problem.solve(Scanner(System.`in`), pw)
    pw.flush()
}

object Problem {
    val MOD = 998244353L
    fun solve(sc: Scanner, pw: java.io.PrintWriter) {
        val N = sc.next().toLong()
        val X = sc.next().toInt()
        val M = sc.next().toInt()
        val S = LongArray(M * 100)
        val Q = LongArray(1000000) { -1L }
        var sum = 0L
        var a = X
        var i = 0L
        while (i < N) {
            if (a == 0) break
            if (Q[a] >= 0) {
                val d = (i - Q[a]).toInt()
                val l = S[i.toInt() - 1] - S[i.toInt() - 1 - d]
                val r = (N - i) / d
                if (r > 0) {
                    sum += l * r
                    i += d * r
                }
                while (i < N) {
                    sum += a
                    a = ((a.toLong() * a.toLong()) % M).toInt()
                    i++
                }
                break
            }
            Q[a] = i

            sum += a
            S[i.toInt()] = sum
            a = ((a.toLong() * a.toLong()) % M).toInt()
            i++
        }
        pw.println(sum)
    }
}