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## 문제

There is a group of $n$ children. According to a proverb, every man to his own taste. So the children value strawberries and raspberries differently. Let us say that $i$-th child rates his attachment to strawberry as $s_{i}$ and his attachment to raspberry as $r_{i}$.

According to another proverb, opposites attract. Surprisingly, those children become friends whose tastes differ.

Let us define friendliness between two children $v$ and $u$ as $$p(u, v) = (s_{u} - s_{v})^{2} + (r_{u} - r_{v})^{2}\text{.}$$

The friendliness between three children $v$, $u$, $w$ is half the sum of pairwise friendlinesses: $$p(u, v, w) = \frac{p(u, v) + p(u, w) + p(v, w)}{2}\text{.}$$

The best friends are such pairs of children $(u, v)$ that $u \ne v$ and $p(u, v) \ge p(u, v, w)$ for every $w$. Your goal is to find all pairs of best friends.

## 입력

In the first line there is one integer $n$, the number of children ($2 \le n \le 2 \cdot 10^{5}$).

Each of the next $n$ lines contains two integers $s_{i}$ and $r_{i}$ ($-10^{8} \le s_{i}, r_{i} \le 10^{8}$).

It is guaranteed that, for every two children, their tastes differ. In other words, if $u \ne v$, then $s_{u} \ne s_{v}$ or $r_{u} \ne r_{v}$.

## 출력

On the first line, output the number of pairs of best friends.

After that, output those pairs. Each pair should be printed on a separate line. A pair is denoted by two integers: the indices of children in this pair. Children are numbered in the order of input starting from $1$. You can output pairs in any order. You can output indices in each pair in any order.

It is guaranteed that the required number of pairs doesn't exceed $10^{6}$.

## 예제 입력 1

4
0 0
1 0
0 1
1 1


## 예제 출력 1

2
1 4
2 3


## 예제 입력 2

3
0 0
0 10
5 8


## 예제 출력 2

0