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>
#define LL long long
const LL MOD = 998244353;
const int MAXN = 100005;
using namespace std;

int N, L, R;
vector<int> G[2*MAXN];
bool vis[2*MAXN];

void dfs(int v) {
	if (v < MAXN) L++;
	else R++;
	
	vis[v] = true;
	for (int to : G[v]) {
		if (!vis[to]) {
			dfs(to);
		}
	}
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	cin >> N;
	for (int i = 0; i < N; i++) {
		int x, y;
		cin >> x >> y;
		x--; y--;
		G[x].push_back(y + MAXN);
		G[y + MAXN].push_back(x);
	}
	
	LL tot = 0;
	for (int i = 0; i < MAXN; i++) {
		L = R = 0;
		dfs(i);
		tot += (LL) L * R;
	}
	
	cout << tot - N << '\n';
}