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
import sys
from collections import deque  # 双方向キュー
from collections import defaultdict  # 初期化済み辞書
from heapq import heapify, heappush, heappop, heappushpop  # プライオリティキュー
from bisect import bisect_left, bisect_right  # 二分探索
#import numpy as np  # 1.8.2
#import scipy  # 0.13.3
sys.setrecursionlimit(int(1e+7))  # スタック上限

s2nn = lambda s: [int(c) for c in s.split(' ')]
ss2nn = lambda ss: [int(s) for s in ss]
ss2nnn = lambda ss: [s2nn(s) for s in ss]
i2s = lambda: sys.stdin.readline().rstrip()
i2n = lambda: int(i2s())
i2nn = lambda: s2nn(i2s())
ii2ss = lambda n: [sys.stdin.readline().rstrip() for _ in range(n)]
ii2nn = lambda n: ss2nn(ii2ss(n))
ii2nnn = lambda n: ss2nnn(ii2ss(n))

def main():
    N = i2n()
    XY = ii2nnn(N)
    M = int(1e+5) + 5

    # 二部グラフ
    to = [[] for _ in range(M*2)]
    cnt = [0] * (M*2)
    visited = [False] * (M*2)

    # 二部グラフを構築
    for x, y in XY:
        y += M
        to[x].append(y)
        to[y].append(x)

    cnt = [0, 0]
    def bfs(n):
        q = deque()
        q.append(n)
        while q:
            i = q.popleft()
            if visited[i]:
                continue
            visited[i] = True
            cnt[i // M] += 1
            for j in to[i]:
                q.append(j)

    # 二部グラフを走査
    ans = 0
    for i in range(M*2):
        if visited[i]:
            continue
        cnt[0] = 0
        cnt[1] = 0
        bfs(i)
        ans += cnt[0] * cnt[1]
    
    ans -= N  # 既にある点は操作数に含まれない
    print(ans)

main()
