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 <cctype>
#define int long long

const int MAXN = 100000;

int n, ans;
int f[200005], a1[200005], a2[200005];

int read() {
    int ret, f = 1;
    char ch;
    while(!isdigit(ch = getchar())) (ch == '-') && (f = -1);
    for(ret = ch - '0'; isdigit(ch = getchar()); ret *= 10, ret += ch - '0');
    return ret * f;
}

void print(int x) {
    if(x < 0) putchar('-'), x = -x;
    if(x > 9) print(x / 10);
    putchar(x % 10 + '0');
}

void init(int x) {
    for(int i = 1; i <= x; ++i) f[i] = i;
}

int find(int x) {
    return f[x] == x ? x : f[x] = find(f[x]);
}

void merge(int x, int y) {
    f[find(x)] = find(y + MAXN);
}

signed main() {
    n = read();
    init(MAXN << 1);
    for(int i = 1; i <= n; ++i)
        merge(read(), read());
    for(int i = 1; i <= MAXN; ++i) ++a1[find(i)];
    for(int i = MAXN + 1; i <= MAXN << 1; ++i) ++a2[find(i)];
    for(int i = 1; i <= MAXN << 1; ++i)
        ans += a1[i] * a2[i];
    return print(ans - n), 0;
}