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)
db = Array.new(Math.log(n,2).to_i+1).map{Array.new(m)}
s = Array.new(Math.log(n,2).to_i+1).map{Array.new(m)}

y = x
while true
  if db[0][y**2%m]
    db[0][y] = y**2%m
    s[0][y] = y**2%m
    break
  else
    db[0][y] = y**2%m
    s[0][y] = y**2%m
    y = y**2%m
  end
end
for i in 0..(Math.log(n,2).to_i-1)
  for j in 0..(m-1)
    if  db[i][j] != nil
      db[i+1][j] = db[i][db[i][j]]
      s[i+1][j] = s[i][j] + s[i][db[i][j]]
    end
  end
end
sum = x
pre = x
for k in 0..(Math.log(n,2))
  if n[k] == 1
    sum += s[k][pre]
    pre = db[k][pre]
  end
end
puts sum-pre