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 Julia (1.4.0) (judge verdict: Accepted)

parseline(str=readline()) = parse.(Int, split(str))
function main()
    N, X, M = parseline()
    A = fill(-1, M+1)
    A[1] = X
    e = 0
    flag = zeros(Int, M+1)
    flag[X+1] = 1
    for m in 2:M+1
        temp = powermod(A[m-1], 2, M)
        if flag[temp+1] > 0
            A[m] = temp
            e = m
            break
        end
        flag[temp+1] = 1
        A[m] = temp
    end
    s = findfirst(x->x==A[e], A)
    ans = s>1 ? sum(A[1:s-1]) : 0
    cycle = A[s:e-1]
    per = length(cycle)
    T = (N-s)÷per
    S = mod(N-s, per)
    ans += T*sum(cycle) + sum(cycle[1:S+1])
    println(ans)
end
main()