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 xi and yi (1 ≤ xi, yi ≤ 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.
Example
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.
**好题，大致意思是判断两点是否在同一个对角线上。看数据很容易就超时，所以不考虑点，而是考虑对角线，考虑在同一个对角线上有几个点，几个点可以构成几组对角线。
首先来说在同一主对角线上，x-y相同，在同一副对角线上x+y相同。那么我们认为加和相同的点和差相同的点视为在同一对角线上，由此可以知道一个对角线上有几个点。
假如有n个点那就有n+n-1+n-2+…..+1个对角线。
结果就显而易见，(n*n-1)/2
**
#include<stdio.h>
int k1[2000+10]={0};
int k2[2000+10]={0};
int main()
{
int n;
scanf("%d",&n);
int x, y;
for(int i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
k1[x+y]++;
k2[x-y+1000]++;
}
long long sum=0;
for(int i=0;i<2000;i++)
{
if(k1[i])
sum+=(k1[i]*(k1[i]-1))/2;
if(k2[i])
sum+=(k2[i]*(k2[i]-1))/2;
}
printf("%d
",sum);
return 0;
}