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 Ruby (2.7.1) (judge verdict: Wrong Answer)

n, x, m = gets.split(" ").map &:to_i
a = []
i = 0
flg = true
aa = 0
bb = 0
while flg
  if a.include?(x)
    aa = i
    bb = x
    flg = false
  else
    a.push(x)
  end
  x = (x%m)**2
  x %= m
  i += 1
  if i > n # || i >= Math.sqrt(m) 鳩ノ巣法
    flg = false
  end
end
if aa == 0
  puts a[0..n-1].sum
  exit
end
if n < aa
  puts a[0..n].sum
  exit
end
f = a.find_index(bb)
Q1 = a[0..f-1].sum
nn = n - f
Q2 = nn/(aa - f) * a[f..aa].sum
Q3 = a[f..(f+n%aa)].sum

#p Q1, Q2, Q3
p Q1 + Q2 + Q3