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 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.

Sample test(s)

input

51 11 53 35 15 5

output

6

input

31 12 33 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.

解题说明：题意是1个1000*1000的地图，上面有一些点，如果两个点在一个对角线上，那么他们就可以互相攻击，求共有多少对这样的点。做法是考虑正副对角线的规律，正对角线上的点i-j相同，副对角线上的点i+j相同，这样这道题就很好求了，一个数组保存各个对角线的规律值，因为i-j会有负数，所以扩大数组，i+j极限值为2000,i-j的最小值为-999，所以说数组极限值要大于2999，最后发现一条线上的对数为一个等差数列，求和就好了。

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<cmath>int main(){int n,i,j,x[2000],y[2000],count=0,a,b;memset(x,0,sizeof(x));memset(y,0,sizeof(y));scanf("%d",&n);for(i=0;i<n;i++){scanf("%d%d",&a,&b);x[a+b-2]++;y[a-b+999]++;}for(i=1;i<1998;i++){if(x[i]){count+=(x[i]*(x[i]-1))/2;}if(y[i]){count+=(y[i]*(y[i]-1))/2;}}printf("%d\n",count);return 0;}