Watchmen （Codeforces Round #345 (Div. 2) C）

C. 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.

题意：找出有多少对点是|xi - xj| + |yi - yj|=





解题思路：只要一个坐标x或y中一个与另一个坐标x或y相等就行，那么搞两个数组，一个按照x排序，

一个按照y排序，找出相同的，按排列组合公式计算。但要注意有可能如果有相同的点，

那么就会多算了一次这些相同的点，故最后减去相同的点的组合数。

/* ***********************************************
┆  ┏┓　　　┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　━　　　┃ ┆
┆┃　┳┛　┗┳　┃ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　┻　　　┃ ┆
┆┗━┓　马　┏━┛ ┆
┆　　┃　勒　┃　　┆　　　　　　
┆　　┃　戈　┗━━━┓ ┆
┆　　┃　壁　　　　　┣┓┆
┆　　┃　的草泥马　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
using namespace std;

#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define maxn 200005
#define f first
#define se second
typedef pair<int,int> p;
vector<p> sx;
vector<p> sy;

int main()
{
int n;
int x,y;
cin>>n;
for(int i=0;i<n;i++)
{

scanf("%d%d",&x,&y);
sx.push_back(p(x,y));
sy.push_back(p(y,x));
}
sort(sx.begin(),sx.end());
sort(sy.begin(),sy.end());
int temp=sx[0].f;
ll cnt=1,sumx=0;
for(int i=1;i<n;i++)
{
if(sx[i].f==temp)
{
cnt++;
}
else
{
sumx+=(cnt*(cnt-1))/2;
cnt=1;
temp=sx[i].f;
}
}
sumx+=(cnt*(cnt-1))/2;
//cout<<sumx<<endl;
cnt=1;
ll sumy=0;
temp=sy[0].f;
for(int i=1;i<n;i++)
{
if(sy[i].f==temp)
{
cnt++;
}
else
{
sumy+=(cnt*(cnt-1))/2;
cnt=1;
temp=sy[i].f;
}
}
sumy+=(cnt*(cnt-1))/2;
//cout<<sumy<<endl;

int temp2,temp1;
temp2=sx[0].se;
temp1=sx[0].f;
cnt=1;
ll sub=0;
for(int i=1;i<n;i++)
{
if(sx[i].f==temp1 && sx[i].se==temp2)
{
cnt++;
}
else
{
sub+=(cnt*(cnt-1))/2;
cnt=1;
temp2=sx[i].se;
temp1=sx[i].f;
}
}
sub+=(cnt*(cnt-1))/2;

cout<<sumx+sumy-sub<<endl;
return 0;
}