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 bisect import bisect_left

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

N = ir()
XY = [lr() for _ in range(N)]
V = list(range(2 * 10**5 + 1)) # 1-indexed X coordinate、Y coordinateの順番

# Aでxの根を求める
def find(A,x):
    p = A[x]
    if p == x: return x
    a = find(A,p)
    A[x] = a
    return a
# Aでxとyの属する集合の併合
def union(A, x, y):
    if find(A,x) > find(A,y):
        bx, by = find(A,y), find(A,x)
    else:
        bx, by = find(A,x), find(A,y)
    A[by] = bx # 根をbxに統一

for x, y in XY:
    y += 10**5
    if find(V, x) != find(V, y):
        union(V, x, y)

set_V = [set() for _ in range(2 * 10**5 + 1)] # 1-indexed
for x, y in XY:
    root = find(V, x)
    set_V[root] |= set([x, y+10**5])

set_V = [x for x in set_V if x]
answer = 0
for z in set_V:
    z = sorted(list(z))
    cnt_x = bisect_left(z, 10 ** 5 + 1)
    cnt_y = len(z) - cnt_x
    answer += cnt_x * cnt_y

answer -= N
print(answer)
# 48