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 <cstdio>
#include <memory.h>
#define N 100001
int m, xp[N], yp[N], f[N];
int xlast[N], ylast[N];
int xbel[N], ybel[N];
int xnum[N] = {0}, ynum[N] = {0};
int fa(int x) {
	return x == f[x] ? x : f[x] = fa(f[x]);
}
inline void merge(int x, int y) {
	if (fa(x) != fa(y)) f[f[x]] = y;
}
int main() {
	#define init(M) memset(M, -1, sizeof(M))
	init(xlast);
	init(ylast);
	init(xbel);
	init(ybel);
	scanf("%d", &m);
	for (int i = 0; i < m; i++) f[i] = i;
	for (int i = 0; i < m; i++) {
		scanf("%d%d", xp+i, yp+i);
		if (xlast[xp[i]] != -1) merge(xlast[xp[i]], i);
		else xlast[xp[i]] = i;
		if (ylast[yp[i]] != -1) merge(ylast[yp[i]], i);
		else ylast[yp[i]] = i;
	}
	for (int i = 0; i < m; i++) {
		xbel[xp[i]] = ybel[yp[i]] = fa(i);
	}
	for (int i = 1; i <= 100000; i++) {
		if (xbel[i] != -1) xnum[xbel[i]]++;
		if (ybel[i] != -1) ynum[ybel[i]]++;
	}
	long long ans = -m;
	for (int i = 0; i < m; i++) {
		ans += 1ll * xnum[i] * ynum[i];
	}
	printf("%lld", ans);
}