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>
#define rep(i, n) for (int i = 0; i < (n); i++)
using ll = long long;
using namespace std;

pair<ll,ll> dfs(int v, vector<vector<int>> &G, vector<bool> &used){
    used[v] = 1;
    ll x = (v < 100100 ? 1 : 0);
    ll y = 1 - x;
    for (auto nv : G[v]) if (!used[nv]) {
        auto p = dfs(nv,G,used);
        x += p.first;
        y += p.second;
    }
    return make_pair(x,y);
}

int main(){
    int n; cin >> n;
    vector<vector<int>> G(200200);
    rep(i,n){
        int x, y; cin >> x >> y;
        G[x].push_back(y+100100);
        G[y+100100].push_back(x);
    }
    vector<bool> used(200200,0);
    ll ans = 0;
    rep(i,200200){
        if(used[i]) continue;
        auto p = dfs(i,G,used);
        if(p.first <= 1 || p.second <= 1) n -= p.first + p.second - 1;
        else ans += p.first * p.second;
    }
    cout << ans - n << endl;
    return 0;
}