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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxd = 1e5+10;
long long ans;
int n,k,m,fa[maxd<<1],a[maxd<<1][2];
int gf(int x)
{
    if(fa[x] == x) return x;
    return fa[x] = gf(fa[x]);
}
int main()
{
    // freopen("a.in","r",stdin);
    // freopen("k.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<(maxd<<1);i++) fa[i] = i;
    for(int i=1;i<=n;i++)
    {
        int x,y; scanf("%d %d",&x,&y);
        fa[gf(x)] = gf(y+maxd);
    }
    for(int i=1;i<(maxd<<1);i++) a[gf(i)][(i >= maxd)]++;
    for(int i=1;i<(maxd<<1);i++) ans += 1ll * a[i][0] * a[i][1];
    printf("%lld",ans - n);
    return 0;
}