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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

from heapq import heappush, heappop
from collections import defaultdict

N, X, D = map(int, input().split())

if D == 0 :
  print(N + 1 - (N + 1) % 2)

else :
  s = X * N + D * N * (N - 1) // 2

  h = defaultdict(lambda: [])
  for i in range(N + 1) :
    a = X * i + D * i * (i - 1) // 2
    b = X * i + D * i * (2 * N - i - 1) // 2
    
    m = 2 * a - s
    M = 2 * b - s
    
    if M < m :
      m, M = M, m
      
    heappush(h[m % (abs(D) * 2)], (m, M))

  D = abs(D)

  ret = 0
  for heap in h.values() :
    u, v = heappop(heap)
    while heap :
      if heap[0][0] <= v :
        v = max(v, heap[0][1])
        heappop(heap)
      else :
        ret += (v - u) // (2 * D) + 1
        u, v = heappop(heap)
        
    ret += (v - u) // (2 * D) + 1

  print(ret)