codeforces 650A Watchmen 【数学】

A. Watchmen

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen
on a plane, the i-th watchman is located at point (xi, yi).

They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen i and j to
be |xi - xj| + |yi - yj|.
Daniel, as an ordinary person, calculates the distance using the formula

.

The success of the operation relies on the number of pairs (i, j) (1 ≤ i < j ≤ n),
such that the distance between watchman i and watchmen j calculated
by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.

Input

The first line of the input contains the single integer n (1 ≤ n ≤ 200 000) —
the number of watchmen.

Each of the following n lines contains two integers xi and yi (|xi|, |yi| ≤ 109).

Some positions may coincide.

Output

Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.

Examples

input

3
1 1
7 5
1 5

output

2

input

6
0 0
0 1
0 2-1 1
0 1
1 1

output

11

Note

In the first sample, the distance between watchman 1 and watchman 2 is
equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and

for
Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor
Manhattan and Daniel will calculate the same distances.

恩，题意大致是，给出n个点的坐标，求这些坐标中横坐标相同的对数和纵坐标相同的对数，但是当横纵坐标都相同时会多计算一遍因此要找到相同的减去。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define maxn 200020
using namespace std;

struct Node
{
int x,y;
};
Node node[maxn];
bool cmp1(Node a,Node b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
bool cmp2(Node a,Node b)
{
return a.y<b.y;
}
int main()
{
__int64 ans;
int n;
while(~scanf("%d",&n))
{
for(int i=0;i<n;++i)
scanf("%d%d",&node[i].x,&node[i].y);
__int64 cnt=1,num=1;
ans=0;
sort(node,node+n,cmp1);
for(int i=0;i<n;++i)
{
int j=i;
cnt=0;
while(j<n&&node[i].x==node[j].x)
j++,cnt++;
//printf("1cnt=%d\n",cnt);
ans+=cnt*(cnt-1)/2;
i=j-1;
}
for(int i=0;i<n;++i)
{
int j=i;
cnt=0;
while(j<n&&node[i].x==node[j].x&&node[j].y==node[i].y)
j++,cnt++;
// printf("2cnt=%d\n",cnt);
ans-=cnt*(cnt-1)/2;
i=j-1;
}
sort(node,node+n,cmp2);
for(int i=0;i<n;++i)
{
int j=i;
cnt=0;
while(j<n&&node[j].y==node[i].y)
j++,cnt++;
//printf("3cnt=%d\n",cnt);
ans+=cnt*(cnt-1)/2;
i=j-1;
}
printf("%I64d\n",ans);
}
return 0;
}