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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define GET_REP(_1, _2, _3, NAME, ...) NAME
#define rep(...) GET_REP(__VA_ARGS__, irep, _rep)(__VA_ARGS__)
#define rep1(...) GET_REP(__VA_ARGS__, irep1, _rep1)(__VA_ARGS__)
#define _rep(i, n) irep (i, 0, n)
#define _rep1(i, n) irep1(i, 1, n)
#define irep(i, a, n) for (int i = a; i < (int)(n); ++i)
#define irep1(i, a, n) for (int i = a; i <= (int)(n); ++i)
#define rrep(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define rrep1(i, n) for (int i = (int)(n); i >= 1; --i)
#define allrep(X, x) for (auto &&X : x)
#define all(x) begin(x), end(x)
#define debug(x) cout << #x " => " << (x) << endl
#ifdef LOCAL
  #include "../../Lib/cout_container.hpp"
#endif
using lint = long long;
constexpr int MOD = (int)1e9 + 7;
constexpr double EPS = 1e-9;
using namespace std;
namespace { struct INIT { INIT() { cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(15); } } INIT; }

int main(void) {
  lint n, x, d;
  cin >> n >> x >> d;
  if (d == 0) {
    if (x == 0) {
      cout << 1 << endl;
    } else {
      cout << n + 1 << endl;
    }
    return 0;
  }
  if (d < 0) {
    x = -x;
    d = -d;
  }
  unordered_map<int, vector<pair<lint, lint>>> now;
  rep1 (A, 0, n) {
    // S = Ax + Bd
    lint Bmin = (A - 1) * A / 2, Bmax = (2 * n - A - 1) * A / 2;
    lint Adiv = A * x / d, Smod = A * x % d;
    if (Smod < 0) {
      Smod += abs(d);
      ++Adiv;
    }
    now[Smod].push_back({Adiv + Bmin, Adiv + Bmax});
  }
  lint cnt = 0;
  allrep (m, now) {
    sort(all(m.second));
    lint now = LLONG_MIN;
    allrep (p, m.second) {
      lint l = max(now, p.first), r = max(now, p.second + 1);
      cnt += r - l;
      now  = r;
    }
  }
  cout << cnt << endl;
  return 0;
}