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
def main():
    import sys
    input = sys.stdin.readline
    sys.setrecursionlimit(10**7)
    from collections import Counter, deque
    from itertools import combinations, permutations, accumulate, groupby, product
    from bisect import bisect_left,bisect_right
    from heapq import heapify, heappop, heappush
    import math
    #from math import gcd

    #inf = 10**17
    #mod = 10**9 + 7

    n = int(input())
    visited_x = [0]*(10**5+1)
    visited_y = [0]*(10**5+1)
    adj_x = [[] for _ in range(10**5+1)]
    adj_y = [[] for _ in range(10**5+1)]

    # 二部グラフにおける隣接点
    for _ in range(n):
        x, y = map(int, input().split())
        adj_x[x].append(y)
        adj_y[y].append(x)

    res = 0
    for x in range(1, 10**5+1):
        if visited_x[x] or not adj_x[x]:
            continue
        visited_x[x] = 1
        vx = 1
        vy = 0
        #[x座標, 軸の方向]
        d = deque([[x, 0]])
        while d:
            v, axis = d.popleft()
            if axis == 0:
                for nv in adj_x[v]:
                    if visited_y[nv]:
                        continue
                    visited_y[nv] = 1
                    vy += 1
                    d.append([nv, 1])
            else:
                for nv in adj_y[v]:
                    if visited_x[nv]:
                        continue
                    visited_x[nv] = 1
                    vx += 1
                    d.append([nv, 0])
        res += vx * vy

    print(res-n)

if __name__ == '__main__':
    main()