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
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda :sys.stdin.readline().rstrip()
def resolve():
    V=100005
    n=int(input())
    E=[[] for _ in range(V*2)]
    for _ in range(n):
        x,y=map(int,input().split())
        y+=V
        E[x].append(y)
        E[y].append(x)

    visited=[0]*(2*V)
    def dfs(v):
        if(visited[v]): return
        visited[v]=1
        cnt[v//V]+=1
        for nv in E[v]: dfs(nv)

    ans=[0]
    for v in range(2*V):
        if(visited[v]): continue
        cnt=[0]*2 # 連結成分のx,yの個数
        dfs(v)
        ans[0]+=cnt[0]*cnt[1]
    print(ans[0]-n)
resolve()