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
from collections import Counter
import sys
input = sys.stdin.readline

# Union-Findデータ構造
class UnionFind:
    def __init__(self, numV):
        self.pars = list(range(numV))
        self.ranks = [0] * numV
    def find(self, x):
        if self.pars[x] == x: return x
        else:
            self.pars[x] = self.find(self.pars[x])
            return self.pars[x]
    def union(self, x, y):
        x, y = self.find(x), self.find(y)
        if x == y: return
        if self.ranks[x] < self.ranks[y]:
            self.pars[x] = y
        else:
            self.pars[y] = x
            if self.ranks[x] == self.ranks[y]:
                self.ranks[x] += 1
    def same(self, x, y):
        return self.find(x) == self.find(y)

N = int(input())
xys = [tuple(map(int, input().split())) for _ in range(N)]

M = 10**5
UF = UnionFind(2*M)
for x, y in xys:
    x, y = x-1, y-1
    UF.union(x, y+M)

for x in range(2*M):
    UF.find(x)

cntX = Counter(UF.pars[:M])
cntY = Counter(UF.pars[M:])

ans = 0
for v in cntX.keys():
    ans += cntX[v] * cntY[v]

print(ans - N)
