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;

const int maxn = 200005;
const int q = 100000;

int n, x, y;
long long ans, sz_x, sz_y, edges;
vector < int > g[maxn];
bool used[maxn];

void dfs (int u) {
	used[u] = 1;
	if (u > q)
		++sz_y;
	else
		++sz_x;
	for (int v: g[u]) {
		++edges;
		if (!used[v])
			dfs(v);
	}
}

signed main () {
	scanf("%d", &n);
	for (int i = 0; i < n; ++i) {
		scanf("%d %d", &x, &y);
		g[x].push_back(q + y);
		g[q + y].push_back(x);
	}
	for (int i = 1; i <= q; ++i) {
		if (!used[i]) {
			sz_x = sz_y = edges = 0;
			dfs(i);
			ans += sz_x * sz_y - edges / 2;
		}
	}
	printf("%lld\n", ans);
}
