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 Python (3.8.2) (judge verdict: Accepted)

a = [int(s) for s in input().split()]
b = a[1]
c = [b]
d = {b:0}
flag = int(0)
k = int(0)
t = int(0)
m = int(0)
suma = int(0)
sumb = int(0)
cnt = int(0)
newc = []
newnewc = []

for i in range(1, a[0]):
    b = (b ** 2) % a[2]
    if d.get(b) == None:
        d[b] = i
        c.append(b)
    else:
        flag = 1
        sumb = sum(c)
        k = int(c.index(b))
        newc = c[k:]
        suma = sum(newc)
        k = len(newc)
        t = i
        m = (a[0] - t) % k
        break

if flag == 1:
    newnewc = newc[:m]
    print((a[0] - t) // k * suma + sumb + sum(newnewc))
else:
    print(sum(c))