Codeforces Round #341 (Div. 2) B

Description

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 ≤ 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.

Sample test(s)

input

5
1 1
1 5
3 3
5 1
5 5

output

6

input

3
1 1
2 3
3 5

output

0
建议不要看下面的NOTE 主要还是说对角线的关系，也就是 x+y x-y当然因为x-y可能是负数，我们加1000来纠正一下
然后一条对角线选两个点。。组合公式套一套就好

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x7fffffff
#define INFL 0x7fffffffffffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
LL a[10000],b[10000];
int main()
{
    int n;
    LL ans=0;
    int x,y;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>x>>y;
        a[x+y]++;
        b[x-y+1000]++;
    }
    for(int i=0;i<=10000;i++)
    {
        ans+=a[i]*(a[i]-1)/2;
        ans+=b[i]*(b[i]-1)/2;
    }
    cout<<ans<<endl;
    return 0;
}