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

def wantp(n):

    q = []
    while p[n] != n:
        q.append(n)
        n = p[n]

    for i in q:
        p[i] = n

    return n

N = int(input())

oXYi = []
XYi = []
YXi = []

for i in range(N):

    x,y = map(int,input().split())
    oXYi.append([x,y,i])
    XYi.append([x,y,i])
    YXi.append([y,x,i])

p = [i for i in range(N)]
rank = [1] * N

XYi.sort()
YXi.sort()


li = -1
for i in range(N):

    if XYi[i][0] != XYi[li][0]:
        li = i

    else:
        ti = XYi[i][2]
        tli = XYi[li][2]
        ip = wantp(ti)
        lip = wantp(tli)

        if ip != lip:
        
            if rank[ip] < rank[lip]:
                p[ip] = lip
            elif rank[ip] > rank[lip]:
                p[lip] = ip
            else:
                p[lip] = ip
                rank[ip] += 1

li = -1
for i in range(N):

    if YXi[i][0] != YXi[li][0]:
        li = i

    else:
        ti = YXi[i][2]
        tli = YXi[li][2]
        ip = wantp(ti)
        lip = wantp(tli)
        
        if ip != lip:
        
            if rank[ip] < rank[lip]:
                p[ip] = lip
            elif rank[ip] > rank[lip]:
                p[lip] = ip
            else:
                p[lip] = ip
                rank[ip] += 1       

ndic = {}
xkazu = {}
ykazu = {}

for i in range(N):

    np = wantp(i)
    if np not in ndic:
        ndic[np] = 1
    else:
        ndic[np] += 1

    if np not in xkazu:
        xkazu[np] = {}
        ykazu[np] = {}

    xkazu[np][oXYi[i][0]] = 1
    ykazu[np][oXYi[i][1]] = 1

ans = 0

for i in ndic:
    ans += len(xkazu[i]) * len(ykazu[i]) - ndic[i]

print (ans)