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 Ruby (2.7.1) (judge verdict: Accepted)

n, x, m = gets.chomp!.split(" ").map(&:to_i)

if x == 0
    puts 0
    exit 0
elsif x == 1
    puts n
    exit 0
end

memo = {}

ans = x
sum = x
memo[ans] = 1

b_i = -1
2.upto(n) do |i|
    ans = (ans ** 2) % m
    sum += ans
    if memo[ans]
        b_i = i - 1
        sum -= ans
        break
    else
        memo[ans] = 1
    end
end

if b_i == -1
    puts sum
else
    # 残回数
    rem = n - memo.length

    j = -1
    found = false
    sub = 0
    memo.each.with_index(0) do |(k, _), i|
        if k == ans
            found = true
            j = i
        end

        if found
            sub += k
        end
    end

    # 1ループ数
    lp = memo.length - j

    sum += ((rem / lp) * sub)
    #pp sum
    rem2 = rem % lp
    #pp rem2
    x = memo.keys
    #pp x
    j.upto(rem2+j-1) do |i|
        sum += x[i]
    end

    puts sum
end