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.chomp.split.map(&:to_i)    # 要素数、初期値、割る数
id = Array.new(m, -1)    # 
a = []
len = tot = 0
while id[x] == -1
  a << x    # 
  id[x] = len   # id[x] に、ｘが何番目にでてきたか入れる
  len += 1    # 何番目か
  tot += x    # tot に、これまでのｘの累計を入れる
  x = (x*x)%m   # 次のx
end
c = len - id[x]   # c 周期の長さ
s = 0   # s １周期の中のｘの和
for i in id[x]..(len - 1) do
  s += a[i]
end
ans = 0
if n <= len   # 周期に入らず終わる場合
  for i in 0...n do
    ans += a[i]
  end
else  # 周期に入る場合
  ans += tot
  n -= len
  ans += s*(n/c)
  n %= c
  for i in 0...n do
    ans += a[id[x]+i]
  end
end
puts ans