C

题目链接：https://vjudge.net/contest/237394#problem/C

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 (x_i, y_i).

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 |x_i - x_j| + |y_i - y_j|. 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 x_i and y_i (|x_i|, |y_i| ≤ 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.

题目大意：输入n，代表有n个点，下面n行代表每个点的坐标，求用以上两种方式求出答案一样的点有多少个

个人思路：第一想法是直接暴力，(这时候还不太清楚怎么大概判断是否会超时),然后很正常的超时了，这里介绍一下怎么大概估计一下是否会超时

1000ms大概能判断10^7的数据，而且是在数据比较水，条件比较好的情况下，反正在这时候就已经很极限了，可能有时候能判断10^8的数据，这就要求后台数据特别水，而且编译环境好的情况下才有可能ac，而这题时限是3000ms,所以充其量估计也就能判3*10^8的数据,所以直接做肯定是超时的，那么下面看一下用

map来做，嗯。。。很快

还有，两个点要满足条件，必须要x相同或者y相同，但要记住减去重复的点

看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
#include<map>
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
const int maxn=2e5+10;
const ll maxa=32050;
const double x=(1.0+sqrt(5.0))/2;
#define INF 0x3f3f3f3f3f3f
//n*(n-1)/2等价于0+1+2+···+n-1
map<ll,ll>mp1;//用于计算有多少个x相同的点
map<ll,ll>mp2;//用于计算有多少个y相同的点
map<pair<ll,ll>,ll>mp;//用于计算有多少个x和y都相同的点
int main()
{
    mp1.clear();
    mp2.clear();
    mp.clear();//清空
    int n;
    ll a,b,ans=0,num=0;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a>>b;
        num+=mp[make_pair(a,b)];//发现map有个好处，就是关键字是什么都可以，这就大大方便了存储
        mp[make_pair(a,b)]++;
        ans+=mp1[a];
        mp1[a]++;
        ans+=mp2[b];
        mp2[b]++;
    }
    cout<<ans-num<<endl;
    return 0;
}