Codeforces--621B--Wet Shark and Bishops（数学）



B. Wet Shark and Bishops

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from 1 to 1000. Rows are numbered from top to bottom, while columns are numbered from left to right.

Wet Shark thinks that two bishops attack each other if they share the same diagonal. Note, that this is the only criteria, so two bishops may attack each other (according to Wet Shark) even if there is another bishop located between them. Now Wet Shark wants to count the number of pairs of bishops that attack each other.

Input

The first line of the input contains n (1 ≤ n ≤ 200 000) — the number of bishops.

Each of next n lines contains two space separated integers x_i and y_i (1 ≤ x_i, y_i ≤ 1000) — the number of row and the number of column where i-th bishop is positioned. It's guaranteed that no two bishops share the same position.

Output

Output one integer — the number of pairs of bishops which attack each other.

Examples

Input

5
1 1
1 5
3 3
5 1
5 5

Output

6

Input

3
1 1
2 3
3 5

Output

0

Note

In the first sample following pairs of bishops attack each other: (1, 3), (1, 5), (2, 3), (2, 4), (3, 4) and (3, 5). Pairs (1, 2), (1, 4), (2, 5) and (4, 5) do not attack each other because they do not share the same diagonal.

每一个点对应的都有两条斜率为1跟-1的直线，我们可以分别用他们的横截距纵截距两条线，因为斜率可以看做已知

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int a[3000],b[3000];
int main()
{
	long long n;
	while(cin>>n)
	{
		if(n==0) break;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		int x,y;
		for(int i=0;i<n;i++)
		{
			cin>>x>>y;
			int ans1=x-y;
			b[1000-ans1]++;
			int ans2=y+x;
			a[ans2-1]++;
		}
		__int64 sum=0;
		for(int i=0;i<=3000;i++)
		{
			sum+=a[i]*(a[i]-1)/2;
			sum+=b[i]*(b[i]-1)/2;
		}
		cout<<sum<<endl;
	}
	return 0;
}