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<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < n; i++)
const int mx = 100005;
int n, a, b;
int p[mx], yp[mx];
long long ans, nmy[mx];
void ufini(int n){ fill(p, p+n, -1);}
int find(int x){return (p[x] < 0 ? x : p[x] = find(p[x]));}
void uni(int x, int y){
    x = find(x); y = find(y);
    if(x != y){
        if(x < y) swap(x, y);
        nmy[x] += nmy[y];
        p[x] += p[y];
        p[y] = x;
    }
}
int main(){
    scanf("%d", &n);
    fill(p, p+mx, -1);
    fill(yp, yp+mx, -1);
    rep(i,n){
        scanf("%d%d", &a, &b);
        if(yp[b] == -1){
            yp[b] = find(a);
            nmy[yp[b]]++;
        }else uni(a, yp[b]);
    }
    rep(i,mx) if(find(i) == i) ans -= nmy[i] * p[i];
    printf("%lld\n", ans - n);
}