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)
ary = [x]
v = Array.new(m)
v[x] = 0

a = x * x % m
i = 0
while !v[a]
    v[a] = (i += 1)
    ary << a
    a = a * a % m
end

if i + 1 >= n
    ans = ary[0, n].sum
else
    j = v[a]
    ans = ary[0, j].sum
    n -= j

    s, t = n.divmod(i - j + 1)
    ans += ary[j .. i].sum * s
    ans += ary[j, t].sum
end

p ans