2017暑期集训——Wet Shark and Bishops(思维）

J - *Wet Shark and Bishops（思维）

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 x_i and y_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.

Example

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.

思路：由于象只能斜着走，即沿着主对角线或者副对角线的方向走，路线与主对角线或者副对角线平行；与主对角线方向平行的路线上的点纵坐标与横坐标之差都相等，与副对角线方向上的平行的路线横坐标和纵坐标之后都相等；用两个数组分别表示横纵坐标的和与差，相等的则在一条对角线上，会互相攻击，同一个点会有两个方向，所以两个运算都要参与；横纵坐标之和为1到2000；之差为-999到999，数组下标不能为负数，所以相减 后加1000，变成1到1999；所以两个数组开到比2000大一点哟；对应于同一个下标，进行数组计数，从n个中进行排列组合选出2个，即Cn(2)=n*(n-1)/2;然后所有的都相加起来。

#include<cstdio>#include<cstring>#define MAX 2000+22int main(){int n;long long a[MAX],b[MAX];while(~scanf("%d",&n)){memset(a,0,sizeof(a));memset(b,0,sizeof(b)); for(int i=0;i<n;i++){int x,y;scanf("%d%d",&x,&y);a[x+y]++;b[y-x+1000]++; }long long sum=0;for(int i=0;i<=2000;i++){if(a[i])sum+=(a[i]*(a[i]-1))/2;if(b[i])sum+=(b[i]*(b[i]-1))/2;}printf("%lld\n",sum);}return 0;}