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: Accepted)

ma = lambda :map(int,input().split())
lma = lambda :list(map(int,input().split()))
tma = lambda :tuple(map(int,input().split()))
ni = lambda:int(input())
yn = lambda fl:print("Yes") if fl else print("No")
ips = lambda:input().split()
import collections
import math
import itertools
import heapq as hq
n,x,m = ma()
idx = [[] for i in range(m)]
tmp=[]
ans = 0
l=0
r=0
for i in range(m+3):
    x = x%m
    tmp.append(x)
    #print(tmp)
    if idx[x]==[]:
        idx[x].append(i)
    else:
        l = idx[x][0]
        r=i
        break
    ans+=x
    x*=x
    if i==n-1:
        print(ans)
        exit()
#print(l,r)
s =sum(tmp[l:r])
d = r-l
p,q = divmod(n-r,d)
ans+=s*p
for i in range(l,l+q):
    ans+=tmp[i]
print(ans)