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;

typedef long long ll;

const int Maxn = 200005;
const int nil = 100001;

int n;
int par[Maxn], siz[Maxn];
int mya[Maxn], myb[Maxn];
ll res;

int getPar(int x) { return par[x] == x? x: par[x] = getPar(par[x]); }

void unionSet(int a, int b)
{
    a = getPar(a), b = getPar(b);
    if (a == b) return;
    if (siz[a] < siz[b]) swap(a, b);
    siz[a] += siz[b]; par[b] = a;
    mya[a] += mya[b]; myb[a] += myb[b];
}

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < nil; i++) {
        par[i] = i; siz[i] = 1;
        mya[i] = 1;
    }
    for (int i = nil; i < Maxn; i++) {
        par[i] = i; siz[i] = 1;
        myb[i] = 1;
    }
    for (int i = 0; i < n; i++) {
        int x, y; scanf("%d %d", &x, &y);
        unionSet(x, nil + y);
    }
    for (int i = 0; i < Maxn; i++) if (getPar(i) == i)
        res += ll(mya[i]) * myb[i];
    res -= n;
    cout << res << endl;
    return 0;
}
