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;

int par[200005];
long long xs[200005];
long long ys[200005];
int getpar(int x){
    if(par[x] != x){
        par[x] = getpar(par[x]);
    }
    return par[x];
}

void join(int x, int y){
    int xp = getpar(x);
    int yp = getpar(y);
    if(xp == yp) return;
    xs[xp] += xs[yp];
    ys[xp] += ys[yp];
    par[yp] = xp;
    xs[yp] = 0;
    ys[yp] = 0;
}

int main(){
    int N;
    scanf("%d", &N);
    for(int i = 1; i <= 100000; i++){
        par[i] = i;
        xs[i] = 1;
        ys[i] = 0;
    }
    for(int i = 100001; i <= 200000; i++){
        par[i] = i;
        xs[i] = 0;
        ys[i] = 1;
    }
    for(int i = 0; i < N; i++){
        int tx, ty;
        scanf("%d %d", &tx, &ty);
        ty += 100000;
        join(tx, ty);
    }
    long long tal = 0;
    for(int i = 1; i <= 200000; i++){
        tal += xs[i] * ys[i];
    }
    long long n = N;
    printf("%lld", tal - n);
}
