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: Accepted)

END { load __FILE__ unless $stdin.eof? }

n, x, m = gets.split.map!(&:to_i)

def f(x, m)
  x * x % m
end

x1 = x2 = x
i = 0
begin
  i += 1
  x1 = f(x1, m)
  x2 = f(f(x2, m), m)
end while x1 != x2

x1 = x
j = 0
begin
  j += 1
  x1 = f(x1, m)
  x2 = f(x2, m)
end while x1 != x2

k = 0
begin
  k += 1
  x1 = f(x1, m)
  x2 = f(f(x2, m), m)
end while x1 != x2

if n <= j
  s = n.pred.times.sum(x) { x = f(x, m) }
  puts s
else
  s0 = j.pred.times.sum(x) { x = f(x, m) }

  n -= j
  s1 = k.times.sum { x = f(x, m) }
  s2 = (n % k).times.sum { x = f(x, m) }

  puts s0 + s1 * (n / k) + s2
end