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 watch(x) cerr << (#x) << ": " << (x) << endl
#define int long long

const int N = 100000;
int n;
vector<int> graph[N*2+1];
int seen[N*2+1];

void dfs(int cur, int& l, int& r) {
    seen[cur] = true;
    if(cur <= N) l++;
    else r++;
    for(auto child : graph[cur]) {
        if(!seen[child]) {
            dfs(child, l, r);
        }
    }
}

int32_t main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n;
    for(int i = 0; i < n; ++i) {
        int x, y; cin >> x >> y;
        graph[x].push_back(y+N);
        graph[y+N].push_back(x);
    }

    int ans = 0;
    for(int i = 1; i <= N; ++i) {
        if(!seen[i]) {
            int l = 0, r = 0;
            dfs(i, l, r);
            ans += l*r;
        }
    }
    ans -= n;

    cout << ans << endl;
    
    return 0;
}
