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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)


n,x,m=map(int,input().split())

if x==0:
    print(0)
elif x==1:
    print(n)
elif m==1:
    print(0)
elif n<=m*2:
    asum=x
    ai=x
    for i in range(1,n):
        ai=(ai*ai)%m
        asum+=ai
    print(asum)
    
else:
    aa=[]
    ai=x
    icnt=0
    asum=0
    while not ai in aa:
        asum+=ai
        aa.append(ai)
        ai=(ai*ai)%m
        icnt+=1
    asum0=0
    ai0=x
    icnt0=0
    for i in range(icnt+1):
        asum0+=ai0
        ai0=(ai0*ai0)%m
        icnt0+=1
        if ai0==ai:
            break
    icntk=icnt-icnt0
    if icntk==0:
#        print("0:",ai,asum,icnt,ai0,asum0,icnt0)
        icntk=1
        n0=n-1
        nk=n0//icntk
        nr=0
        asumk=ai0
        asumr=0
        ans=x*n
        print(ans)       
    else:
#        print("1:",ai,asum,icnt,ai0,asum0,icnt0)
        n0=n-1-icnt0
        nk=n0//icntk
        nr=n0%icntk
        asumk=asum-asum0
        asumr=0
        air=ai
        for i in range(nr):
            asumr+=air
            air=(air*air)%m
        asumr+=air
       
        ans=asum0+asumk*nk+asumr    
        print(ans)