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
N = int(input())
x, y = (
    zip(*(map(int, input().split()) for _ in range(N))) if N else
    ((), ())
)


class UnionFindTree:
    def __init__(self, n):
        self.p = [i for i in range(n + 1)]
        self.r = [1 for _ in range(n + 1)]

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, x, y):
        px = self.find(x)
        py = self.find(y)
        if px != py:
            if self.r[px] < self.r[py]:
                self.p[px] = py
                self.r[py] += self.r[px]
            else:
                self.p[py] = px
                self.r[px] += self.r[py]

    def same(self, x, y):
        return self.find(x) == self.find(y)


uft = UnionFindTree(10**5)
X = [[] for _ in range(10**5 + 1)]
for a, b in zip(x, y):
    X[a].append(b)
Y = [[] for j in range(10**5 + 1)]
for a, b in zip(x, y):
    Y[b].append(a)
SX = set(x)
SY = set(y)
for j in SY:
    for i in Y[j]:
        uft.union(Y[j][0], i)

X2 = [[] for _ in range(10**5 + 1)]
for i in SX:
    X2[uft.find(i)].append(i)
ans = 0
for i in SX:
    S = set()
    for i2 in X2[i]:
        for j in X[i2]:
            S.add(j)
    ans += len(X2[i]) * len(S)
ans -= N

print(ans)
