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: Runtime Error)

function ABC_E()
    readints()=parse.(Int,split(readline()))
    N,X,M=readints()
    useS=zeros(Int,M)
    A=zeros(Int,M)
    A[1]=X
    useS[X]=1
    s=X

    for i in 2:M
        X=X*X%M
        X==0 && (break)
        A[i]=X
        s+=X
        if useS[X] > 0
            Ls=useS[X]
            Le=i
            s+=(s-sum(A[1:Ls]))*((N-i)÷(Le-Ls))
            s+=sum(A[Ls:(Ls+(N-i)%(Le-Ls))])
            break
        else
            useS[X]=i
        end
    end
    s
end
print(ABC_E())