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
# https://drken1215.hatenablog.com/entry/2019/06/23/004700
# https://tjkendev.github.io/procon-library/python/union_find/union_find.html

def main():
    import sys
    input = sys.stdin.readline

    MX = 10 ** 5 + 10
    
    *p, = [r for r in range(MX * 2)]

    def root(x):
        if x == p[x]:
            return x
        p[x] = y = root(p[x])
        return y

    def unite(x, y):
        px = root(x)
        py = root(y)
        if px == py:
            return 0
        if px < py:
            p[py] = px
        else:
            p[px] = py
        return 1

    n = int(input())
    for _ in range(n):
        x, y = (int(x) - 1 for x in input().split())  # [1,)->[0,)
        unite(x, y + MX)

    xs = [0] * (MX * 2)
    ys = [0] * (MX * 2)

    for r in range(MX):
        xs[root(r)] += 1

    for r in range(MX, MX * 2):
        ys[root(r)] += 1

    ans = -n
    for r in range(MX * 2):
        ans += xs[r] * ys[r]

    print(ans)


if __name__ == '__main__':
    main()
