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
import sys, math
from collections import defaultdict, deque, Counter
from bisect import bisect_left, bisect_right
from itertools import combinations, permutations, product
from heapq import heappush, heappop
from functools import lru_cache
input = sys.stdin.readline
rs = lambda: input().strip()
ri = lambda: int(input())
rl = lambda: list(map(int, input().split()))
mod = 1000000007
sys.setrecursionlimit(1000000)

N = ri()
to = defaultdict(list)
V = 100005
X = []
for i in range(N):
	x, y = rl()
	X.append(x)
	y += V
	to[x].append(y)
	to[y].append(x)

visited = set()
nx, ny = 0, 0
def dfs(n):
	global nx, ny
	if n in visited: return
	visited.add(n)
	if n < V:
		nx += 1
	else:
		ny += 1
	for m in to[n]:
		dfs(m)

ans = 0
for x in X:
	if x in visited: continue
	nx, ny = 0, 0
	dfs(x)
	ans += nx*ny
print(ans-N)

