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


def main():
    class UnionFind:
        def __init__(self, size):
            self.data = [-1] * size

        def find(self, x):
            if self.data[x] < 0:
                return x
            else:
                self.data[x] = self.find(self.data[x])
                return self.data[x]

        def union(self, x, y):
            x, y = self.find(x), self.find(y)
            if x != y:
                if self.data[y] < self.data[x]:
                    x, y = y, x
                self.data[x] += self.data[y]
                self.data[y] = x
            return (x != y)

    MAX = 10 ** 5 + 1
    uf = UnionFind(2 * MAX)

    N, *XY = map(int, open(0).read().split())

    for x, y in zip(*[iter(XY)] * 2):
        uf.union(x, y + MAX)

    X = Counter(uf.find(i) for i in range(MAX))
    Y = Counter(uf.find(i) for i in range(MAX, MAX * 2))
    res = sum(X[i] * Y[i] for i in range(MAX * 2))

    print(res - N)


if __name__ == '__main__':
    main()
