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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    ll n,x,d;
    cin>>n>>x>>d;
    unordered_map<ll,vector<pair<ll,ll> > >rd;
    for(ll k=0;k<=n;k++){
        ll low=x*k+d*(k*(k-1)/2);
        ll up=x*k+d*((n-1)*k-k*(k-1)/2);
        if(low<=up){
            rd[low%d].push_back(make_pair(low/d-1,up/d));
        }
    }
    ll ans=0;
    for(auto it:rd){
        vector<pair<ll,ll> >vec=it.second;
        sort(vec.begin(),vec.end());
        ll r=-1e18;
        for(auto v:vec){
            ans+=max(0LL,v.second-max(r,v.first));
            r=max(r,v.second);
        }
    }
    cout<<ans<<"\n";
}