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.

根據 |xi - xj| + |yi - yj|. =   

可以推出（xi - xj）*（yi - yj）=0 

所以x相同或者y相同就滿足條件

用map儲存資料 以xi（ yj）為鍵值計算相同的個數

遍歷map 假設於每一個xi（ yj）的值為v    ans+=v*(v-1)/2;(在v個點中任選兩個點）

然後要注意輸入的資料有重複的點 要把這些重複點剪掉ans-=v*(v-1)/2;(在v個點中任選兩個點）

最後要注意精度問題 在計算答案的過程中會爆int

效率 O(nlog(n)+3n)

程式碼如下

 // *******************************************************
// author : Anjone
// source : http://codeforces.com/problemset/problem/650/A
// time   : 2016-4-5 19:28:24
// *******************************************************

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<algorithm>
using namespace std;

#define MAXN 200005
#define Min(_,__) _<__?_:__
#define rep(_,__,___) for(int _=__;_<___;_++)
#define S(_) scanf("%d",&_);
#define P(_) printf("%d\n",_);
#define Pl(_) printf("%I64d\n",_);
#define LL long long

int n , cnt=1;
LL ans = 0;
map<int,int> mx,my;

struct WATCHMEN{
    int x,y;
}w[MAXN];

bool cmp(WATCHMEN a, WATCHMEN b){
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}

int main()
{
    S(n)
    rep(i,0,n){
        scanf("%d%d",&w[i].x,&w[i].y);
        mx[w[i].x]++;
        my[w[i].y]++;
    }
    sort(w,w+n,cmp);
    for(map<int,int>::iterator it = mx.begin() ; it!=mx.end() ; it++){
        if((*it).second>1){
            ans += 1LL*(*it).second*((*it).second-1)/2;
        }
    }
    for(map<int,int>::iterator it = my.begin() ; it!=my.end() ; it++){
        if((*it).second>1){
            ans += 1LL*(*it).second*((*it).second-1)/2;
        }
    }
    rep(i,1,n){
        if(w[i].x==w[i-1].x&&w[i].y==w[i-1].y) cnt++;
        else { ans-=1LL*cnt*(cnt-1)/2 ; cnt = 1;}
    }
    ans-=1LL*cnt*(cnt-1)/2 ;
    Pl(ans)
}