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
#include <bits/stdc++.h>
using namespace std;
#define N 100000

int n, f[N * 2 + 5], cx[N * 2 + 5], cy[N * 2 + 5];

int find(int x) {
    return x == f[x] ? x : f[x] = find(f[x]);
}

int main() {
#ifdef whyqx
    freopen("work.in", "r", stdin);
    freopen("work.out", "w", stdout);
#endif
    cin >> n;
    for (int i = 1; i <= N << 1; ++i)
        f[i] = i;
    for (int i = 1; i <= n; ++i) {
        int x, y;
        scanf("%d%d", &x, &y);
        f[find(x)] = find(y + N);
    }
    for (int i = N + 1; i <= N << 1; ++i)
        cy[find(i)]++;
    for (int i = 1; i <= N; ++i)
        cx[find(i)]++;
    long long ans = 0;
    for (int i = 1; i <= N << 1; ++i)
        ans += 1ll * cx[i] * cy[i];
    cout << ans - n;
    return 0;
}
