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
input = sys.stdin.readline
from collections import *

def bfs(s):
    color[s] = 1
    q = deque([s])
    b, w = 1, 0
    
    while q:
        v = q.popleft()
        
        for nv in G[v]:
            if color[nv]==0:
                color[nv] = -color[v]
                q.append(nv)
                
                if color[nv]==1:
                    b += 1
                else:
                    w += 1
    
    return b*w
                    
N = int(input())
xy, xs, ys = [], [], []

for _ in range(N):
    x, y = map(int, input().split())
    xy.append((x, y))
    xs.append(x)
    ys.append(y)

xs = list(set(xs))
ys = list(set(ys))
xs.sort()
ys.sort()
idx1 = defaultdict(int)

for i in range(len(xs)):
    idx1[xs[i]] = i

idx2 = defaultdict(int)

for i in range(len(ys)):
    idx2[ys[i]] = i+len(xs)

M = len(xs)+len(ys)
G = [[] for _ in range(M)]

for x, y in xy:
    G[idx1[x]].append(idx2[y])
    G[idx2[y]].append(idx1[x])

color = [0]*M
S = 0

for i in range(M):
    if color[i]==0:
        S += bfs(i)

print(S-N)