Wet Shark and Bishops



 Wet Shark and Bishops

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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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 containsn (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 wherei-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 Input

Input

51 11 53 35 15 5

Output

6

Input

31 12 33 5

Output

0

Sample Output

Hint

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.

题意：一个1000*1000的棋盘，给你N个点的坐标，问有多少对点在对角线上。

思路：找出对角线上有多少点，再求C（n,2)即可。

规律：斜向右下的一条|x-y|都相等，斜向左下的一条x+y都相等，求出线上个数。注：相同结果存入一个a[i]，个数为a[i]++,

代码如下：

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int a[2100],b[2100];int main() {   int x,y,n;  while(scanf("%d",&n)!=EOF)  {  memset(a,0,sizeof(a));  memset(b,0,sizeof(b));  long long ans=0;  for(int i=1;i<=n;i++)  {   scanf("%d%d",&x,&y);   a[x-y+1000]++;   b[x+y]++;  }  for(int i=1;i<=2000;i++)  {   if(a[i]!=0)   {    ans+=a[i]*(a[i]-1)/2;   }   if(b[i]!=0)   {    ans+=b[i]*(b[i]-1)/2;   }  }  printf("%lld\n",ans); } return 0;}