There are N dots in a two-dimensional plane. The coordinates of the i-th dot are (x_i, y_i).

We will repeat the following operation as long as possible:

* Choose four integers a, b, c, d (a \neq c, b \neq d) such that there are dots at exactly three of the positions (a, b), (a, d), (c, b) and (c, d), and add a dot at the remaining position.



We can prove that we can only do this operation a finite number of times. Find the maximum number of times we can do the operation.

Constraints

* 1 \leq N \leq 10^5
* 1 \leq x_i, y_i \leq 10^5
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
x_1 y_1
:
x_N y_N


Output

Print the maximum number of times we can do the operation.

Examples

Input

3
1 1
5 1
5 5


Output

1


Input

2
10 10
20 20


Output

0


Input

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5


Output

16
#include <iostream>
#include <vector>
using namespace std;
vector< vector<int> > edg;
vector<bool> vsd;
const int s = 1e5;
void f(int v, int *xcnt, int *ycnt){
    if(vsd[v])return;
    vsd[v] = true;
    if(v < s) (*xcnt)++;
    else (*ycnt)++;
    for(int i = 0; i < edg[v].size(); i++){
        f(edg[v][i], xcnt, ycnt);
    }
}
int main(void){
    int n, xcnt, ycnt;
    long long ans = 0;
    cin >> n;
    edg.assign(s * 2, vector<int>());
    vsd.assign(s * 2, false);

    for(int i=0;i<n;i++){
        int x, y;
        cin >> x >> y;
        x--;
        y += s-1;
        edg[x].push_back(y);
        edg[y].push_back(x);
    }

    for(int i = 0; i < s * 2; i++){
        if(!vsd[i]){
            xcnt = ycnt = 0;
            f(i, &xcnt, &ycnt);
            ans += (long long) xcnt * ycnt;
        }
    }
    cout << ans - n << endl;
}
