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;
#define rep(i, a, b) for(int i = a; i < b; i++)


int N;
ll cnt, cntx, cnty;
vector<bool> used(200000, 0);
vector<vector<int> > G(200000 + 10);

void dfs(int idx, int par){
    used[idx] = true;
    if(idx < 100000) cntx++;
    else cnty++;

    for(auto& to: G[idx]){
        if(to == par) continue;
        cnt++;
        if(used[to]) continue;
        dfs(to, idx);
    }
    return;
}

int main(){
    cin >> N;
    rep(i, 0, N){
        int x, y; cin >> x >> y; x--, y--;
        G[x].push_back(y + 100000);
        G[y + 100000].push_back(x);
    }
    ll ans = 0;
    rep(i, 0, 200000) if(!used[i]){
        cnt = 0;
        cnty =0;
        cntx = 0;
        dfs(i, -1);
        ans += cntx * cnty;
    }
    cout << ans - N << endl;

    return 0;
}