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
N = int(input())
xy = [tuple(map(int,input().split())) for _ in range(N)]
shift = 10**5+1

nodes = {}
visited = {}

for x,y in xy:
    y += shift
    if x in nodes:
        nodes[x].append(y)
    else:
        nodes[x] = [y]
    if y in nodes:
        nodes[y].append(x)
    else:
        nodes[y] = [x]
    visited[x] = False
    visited[y] = False

ans = 0

def dfs(s, nodes,visited):
    q = [s]
    cnt_x = 0
    cnt_y = 0
    
    while q:
        v = q.pop()
        if visited[v]:
            continue
        visited[v] = True
        
        if v//shift < 1:
            cnt_x += 1
        else:
            cnt_y += 1
            
        q.extend(nodes[v])
        
    return visited, cnt_x*cnt_y

for s in nodes:
    if visited[s]:
        continue
    visited, cnt = dfs(s,nodes, visited)
    ans += cnt
    
print(ans - N)