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)

n, x, m = gets.split.map(&:to_i)
a = Array.new(m, nil)
flag = Array.new(m, false)
accum = Array.new(m, 0)
prev = 0
j = n
for i in 0...m do
  a[i] = x
  accum[i] = prev + x
  prev = accum[i]
#  puts "#{a[i]} #{accum[i]}"
  flag[x] = true
  y = x ** 2 % m
  if flag[y]
    j = a.index(y)
#    puts "! #{j}"
    break
  end
  x = y
end
pre_len = j
loop_len = i - j + 1
post_len = ((n - j) % loop_len)
ans = accum[j - 1]
ans += (accum[i] - accum[j - 1]) * ((n - j) / loop_len)
ans += (accum[j + post_len - 1] - accum[j - 1])
puts ans