CodeForces 621B

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 integersx_i andy_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

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.

1000*1000的格子，给出n个点，在对角线上会相互攻击

求会相互攻击的点对数

在平行于主对角线的一条线上，行列差相等

在平行于副对角线的一条线上，行列和相等

利用此规律，数学推一下，去重得结果

#include<iostream>#include<cstdio>#include<string.h>using namespace std;int xay[2005],ysx[2005];int main(){    int n;    int x,y;    while(scanf("%d",&n)!=EOF)    {        long long cnt=0;        memset(xay,0,sizeof(xay));        memset(ysx,0,sizeof(ysx));        int i;        for(i=0;i<n;i++)        {             scanf("%d%d",&x,&y);             xay[x+y]++;             ysx[y-x+1000]++;        }        for(i=0;i<2005;i++)        {            if(xay[i]!=0)                cnt+=(xay[i]*xay[i]-xay[i])/2;        }        for(i=0;i<2005;i++)        {            if(ysx[i]!=0)                cnt+=(ysx[i]*ysx[i]-ysx[i])/2;        }        printf("%lld\n",cnt);    }    return 0;}