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 int long long

const int N = 1e5 + 100;

vector <int> X[N], Y[N];
int x[N], y[N], par[N];
set <int> s[N], a, b;
int n, ans;

int getPar(int v) {
	return (par[v] == v ? v : par[v] = getPar(par[v]));
}

void merge (int v, int u) {
    v = getPar(v);
    u = getPar(u);
	par[u] = v;
}

int32_t main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> x[i] >> y[i];
		X[x[i]].push_back(i);
		Y[y[i]].push_back(i);
		par[i] = i;
	}
	for (int i = 1; i < N; i++) {
		for (auto v : X[i])
			merge(X[i][0], v);
		for (auto v : Y[i])
			merge(Y[i][0], v);
	}
	for (int i = 0; i < n; i++)
		s[getPar(i)].insert(i);
	for (int i = 0; i < n; i++) {
		if (!s[i].size()) continue;
		for (auto v : s[i]) {
			a.insert(x[v]);
			b.insert(y[v]);
	    }
		ans += a.size() * b.size();
		a.clear(); b.clear();
	}
	return cout << ans - n, 0;
}