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)

fun main(args: Array<String>) {
    val sp = readLine()!!.split(" ")

    val n = sp[0].toLong()
    val x = sp[1].toInt()
    val m = sp[2].toInt()


    val first = Array(m) { -1 }
    val sums = Array(m + 1) { 0L }

    var step = 0
    first[x] = step

    var a = x.toLong()
    val period: Int
    val start: Int
    while (true) {
        step++
        sums[step] = (sums[step - 1] + a)
        a = (a * a) % m
        val prev = first[a.toInt()]
        if (prev == -1) first[a.toInt()] = step
        else {
            period = step - prev
            start = prev
            break
        }
    }

    if (n < start) println(sums[n.toInt()])
    else {
        var res = sums[start]
        res += (sums[start + period] - sums[start]) * ((n - start) / period)
        val rest = ((n - start) % period).toInt()
        res +=  sums[start + rest] - sums[start]

        println(res)
    }
}