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 Julia (0.5.0) (judge verdict: Time Limit Exceeded)

function main()
	N,X,D=map(x->parse(Int,x),split(readline()))
	if D==0
		println(X==0 ? 1 : N+1)
		exit()
	end
	if D<0
		D=-D
		X=-X
	end
	S=Pair{Int,Pair{Int,Int}}[]
	for k=0:N
		L=fld(k*X,D)
		U=L+div(k*(k-1),2)=>L+N*k-div(k*(k+1),2)
		push!(S,mod(k*X,D)=>U)
	end
	ans=0
	pm=-1
	pre=0
	for (k,(l,r))=sort(S)
		if k==pm
			ans+=max(0,r-max(pre,l-1))
			pre=max(pre,r)
		else
			pm=k
			ans+=r-l+1
			pre=r
		end
	end
	println(ans)
end
main()