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
# 長方形の4点のうち3点が必要なので自明からほど遠い
# x,yを2部グラフ、格子を辺とみなす
# 連結成分ごとに張られる......
n = int(input())
xy = [list(map(int, input().split())) for _ in range(n)]

mx =0
my = 0
for i in range(n):
    mx = max(mx,xy[i][0])
    my = max(my,xy[i][1])

*p, = range(mx+my+1)
rank = [1]*(mx+my+1)

for i in range(n):
    xy[i][1] += mx

def root(x):
    if x == p[x]:
        return x
    p[x] = y = root(p[x])
    return y
def unite(x, y):
    px = root(x); py = root(y)
    if px == py:
        return 0
    rx = rank[px]; ry = rank[py]
    if ry < rx:
        p[py] = px
    elif rx < ry:
        p[px] = py
    else:
        p[py] = px
        rank[px] += 1
    return 1

for x,y in xy:
    unite (x,y)

xp = [0]*(mx+my+1)
yp = [0]*(mx+my+1)

for i in range(mx+my+1):
    if i<=mx:
        xp[root(i)]+=1#p[i]では不十分
    else:
        yp[root(i)]+=1
ans = 0
for i in range(mx+my+1):
    ans += xp[i]*yp[i]
ans -= n
print(ans)
