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 namespace std;
using ll = long long;
using P = pair<int,int>;

const int MAXN = 100010;
vector<vector<int>> edge(MAXN * 2, vector<int>());
vector<int> visited(MAXN * 2);
ll cntx = 0;
ll cnty = 0;

void dfs(int n){
    if (visited[n]) return;
    visited[n] = 1;
    if (n / MAXN) cnty++;
    else cntx++;
    for(auto nv : edge[n]) {
        dfs(nv);
    }
}

int main(){
    int N; cin >> N;
    rep(i, N){
        int x, y; cin >> x >> y;
        x--;y--;
        edge[x].push_back(y+MAXN);
        edge[y+MAXN].push_back(x);
    }
    ll ans = 0;
    rep(i, MAXN*2){
        if(visited[i]) continue;
        // 連結成分の個数を数える
        cntx = cnty = 0;
        dfs(i);
        ans += cnty * cntx;
    }
    cout << ans - N << endl;
    return 0;
}
