CodeForces 651C

Description

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.

Sample Input

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

Hint

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.

思路_{：把题目的两个公式同时平方，化简得只要xi=xj或者yi=yj，等式就会成立，暴力的话容易超时，学长当时给了题解，}

_{用stl的map去写，后来自己也看了下，的确很好用}

统计x相等的个数加上y相等的个数减去xy同时相等的个数，就是结果了

代码是学长给的，发现比网上的一些简单多了

 #include <iostream>
 #include <map>
 using namespace std;
 int main()
 {
     int n;
     while (cin>>n)
     {
         map<int,int> mx;
         map<int,int> my;
         map<pair<int,int>,int>mxy;
         long long ans = ;
         for (int i=;i<n;++i)
         {
             int x,y;
             cin>>x>>y;
             ans += mx[x]++;
             ans += my[y]++;
             ans -= mxy[make_pair(x,y)]++;
         }
         cout<<ans<<endl;
     }
 }