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(10 ** 7)

def dfs(G, F, crr, pre, ret, st):
    F[crr] = False
    if crr < 10**5:
        ret[0] += 1
    else:
        ret[1] += 1
    for nxt in G[crr]:
        if nxt == pre:
            continue
        p, q = crr, nxt
        if q < p:
            p, q = q, p
        if not (p, q) in st:
            st.add((p, q))
            ret[2] += 1
        if F[nxt]:
            dfs(G, F, nxt, crr, ret, st)

def main():
    n = int(input())
    G = [[] for _ in range(2 * 10**5)]
    for _ in range(n):
        x, y = map(int, input().split())
        x, y = x-1, y-1 + 10**5
        G[x].append(y)
        G[y].append(x)
    F = [True]*(2* 10**5)
    ans = 0
    st = set()
    for i in range(2 * 10**5):
        if F[i] and len(G[i]) > 0:
            tmp = [0]*3
            dfs(G, F, i, -1, tmp, st)
            ans += tmp[0] * tmp[1] - tmp[2]
    print(ans)

if __name__ == "__main__":
    main()