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: Accepted)

INF = 10**30
def solve(n, x, d):
  if d == 0:
    if x == 0:
      return 1
    else:
      return n+1
  D={}
  for k in range(n+1):
    l = k*x + (k-1)*k//2 * d
    r = k*x + (n*k - k*(k+1)//2)*d
    c = k*x % d
    if not c in D:
      D[c] = [];
    l = (l-c)//d
    r = (r-c)//d
    if l > r:
      l, r = r, l
    D[c].append((l, r))
  res = 0
  for v in D.values():
    threshold = -INF
    for l,r in sorted(v):
      l = max(l, threshold)
      res += max(0, r-l+1)
      threshold = max(threshold, r+1)
  return res

n, x, d = map(int, input().split())
print(solve(n, x, d))