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: Time Limit Exceeded)

#include<bits/stdc++.h>

using namespace std;

#define int long long

main() {
#ifdef HOME
    //freopen("input.txt", "r", stdin);
#endif // HOME
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, x, d, ans = 0, sum;
    cin >> n >> x >> d;
    sum = n * (n - 1) / 2;
    set < int > answer;
    for (int k = 0; k <= n; ++k) {
        int imin = k * (k - 1) / 2, imax = n * (n - 1) / 2 - (n - k) * (n - k - 1) / 2;
        //cout << imin << " " << imax << '\n';
        for (int j = 2 * imin - sum; j <= 2 * imax - sum; j++) {
            int xx = (sum + j);
            if (abs(xx) % 2 == 1) continue;
            if (abs(xx / 2) > sum) continue;
            answer.insert((2 * k - n) * x + j * d);
        }
    }
    //for (auto key : answer) cout << key << '\n';
    cout << answer.size() << '\n';
    return 0;
}