Problem:
Score : 600 points

Problem Statement

We have an integer sequence A of length N, where A_1 = X, A_{i+1} = A_i + D (1 \leq i < N ) holds.

Takahashi will take some (possibly all or none) of the elements in this sequence, and Aoki will take all of the others.

Let S and T be the sum of the numbers taken by Takahashi and Aoki, respectively. How many possible values of S - T are there?

Constraints

-10^8 \leq X, D \leq 10^8

1 \leq N \leq 2 \times 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X D

Output

Print the number of possible values of S - T.

Sample Input 1

3 4 2

Sample Output 1

8

A is (4, 6, 8).

There are eight ways for (Takahashi, Aoki) to take the elements: ((), (4, 6, 8)), ((4), (6, 8)), ((6), (4, 8)), ((8), (4, 6))), ((4, 6), (8))), ((4, 8), (6))), ((6, 8), (4))), and ((4, 6, 8), ()).

The values of S - T in these ways are -18, -10, -6, -2, 2, 6, 10, and 18, respectively, so there are eight possible values of S - T.

Sample Input 2

2 3 -3

Sample Output 2

2

A is (3, 0). There are two possible values of S - T: -3 and 3.

Sample Input 3

100 14 20

Sample Output 3

49805

Solution in Ruby (2.3.3) (judge verdict: Accepted)

N, X, D = gets.split.map(&:to_i)

(puts X == 0 ? 1 : N + 1; exit) if D == 0
D < 0 ? (x, d = X, D) : (x, d = -X, -D)
h = Hash.new{|h, k| h[k] = []}
(0 .. N).each do |i|
  l = x*i + i*(i - 1)/2 * d
  r = x*i + i*(N - i + N - 1)/2 * d
  h[l % d] << [l/d, r/d + 1]
end
ans = 0
h.values.each do |a|
  lst = -10**18
  a.sort.each do |s, e|
    if lst < s
      ans += e - s
      lst = e
    end
    if lst < e
      ans += e - lst
      lst = e
    end
  end
end
puts ans