Problem:
Score : 300 points

Problem Statement

There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).

We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:

Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.

Find the minimum total cost required to collect all the balls when we optimally choose p and q.

Constraints

1 \leq N \leq 50

|x_i|, |y_i| \leq 10^9

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 minimum total cost required to collect all the balls.

Sample Input 1

2
1 1
2 2

Sample Output 1

1

If we choose p = 1, q = 1, we can collect all the balls at a cost of 1 by collecting them in the order (1, 1), (2, 2).

Sample Input 2

3
1 4
4 6
7 8

Sample Output 2

1

If we choose p = -3, q = -2, we can collect all the balls at a cost of 1 by collecting them in the order (7, 8), (4, 6), (1, 4).

Sample Input 3

4
1 1
1 2
2 1
2 2

Sample Output 3

2

Solution in Rust (1.42.0) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_imports)]
use ascii::AsciiChar;
use ascii::AsciiStr;
use itertools::Itertools as _;
use permutohedron::LexicalPermutation as _;
use petgraph::algo::dijkstra;
use petgraph::algo::min_spanning_tree;
use petgraph::data::FromElements;
use petgraph::graph::NodeIndex;
use petgraph::graph::UnGraph;
use petgraph::unionfind::UnionFind;
use proconio::fastout;
use proconio::input;
use proconio::marker::Chars;
use proconio::marker::Usize1;
use std::cmp::max;
use std::cmp::min;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;
use superslice::Ext as _;

#[fastout]
fn main() {
    input! {
        N: usize,
        mut XYs: [(i64, i64); N],
    }

    // 扱いやすいように、Xの昇順,Yの昇順でソートしておく
    XYs.sort();

    // Nが小さいので、N^4くらいまで回せる
    // なので、ある2点間についてのxの差やyの差はメモリに乗せることができる

    // 等差数列になるように点を巡っていく際に、-方向に行ける道は逆から辿ることもできる。
    // よって、+方向だけ考えれば充分

    let mut ans = N;

    for fst_ind in 0..N {
        let (fst_x, fst_y) = XYs[fst_ind];

        for snd_ind in fst_ind + 1..N {
            let (snd_x, snd_y) = XYs[snd_ind];
            let p = snd_x - fst_x;
            let q = snd_y - fst_y;

            // pとqが確定していれば、0ポイントで移動できる点は列挙できる
            let mut pq_pair = 0;
            for i in 0..N {
                let (x0, y0) = XYs[i];

                for j in fst_ind + 1..N {
                    let (x1, y1) = XYs[j];

                    if x1 - x0 == p && y1 - y0 == q {
                        pq_pair += 1;
                    }
                }
            }

            let cand = N - pq_pair;
            ans = min(ans, cand);
        }
    }

    println!("{}", ans);
}