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: Runtime Error)

N, X, M = map(int, input().split())

rec = []
s = set()
rec.append(X)
s.add(X)
r = X
idx == -1
idx1 = -1
for i in range(N+1):
    r = (r**2)%M
    if(r not in s):
        rec.append(r)
        s.add(r)
    else:
        rec.append(r)
        idx = i+1
        break
    
for i in range(len(rec)):
    if(rec[i] == rec[idx]):
        idx1 = i
        break

if(idx == -1 and idx1 == -1):
    print(sum(rec))
elif(idx1 != idx):       
    ans = sum(rec[:idx1])
    ans += sum(rec[idx1:idx])*((N-idx1)//(idx-idx1))+sum(rec[idx1:(idx1+(N-idx1)%(idx-idx1))])
    print(ans)
else:
    ans = sum(rec[:idx1])
    ans += rec[idx]*(N-idx1)
    print(ans)