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 JavaScript (node.js v5.12) (judge verdict: Accepted)

"use strict"
function main(arg) {
    arg = arg.trim().split("\n")
    var N = Number(arg.shift())
    var data = arg.map((e1) => {
        return e1.split(" ").map(Number)
    })
    var pq = getPq(data)
    var p = pq[0]
    var q = pq[1]
    var tmp = []
    for (var i = 0; i < data.length; i++) {
        for (var j = 0; j < data.length; j++) {
            if (i === j) continue
            var _p = data[i][0] - data[j][0]
            var _q = data[i][1] - data[j][1]
            if (p === _p && q === _q) {
                // console.log(i)
                tmp.push(i)
            }
        }
    }
    var b = tmp.filter(function (x, i, self) {
        return self.indexOf(x) === i;
    });
    console.log(data.length - b.length)
    // console.log(pq)
    return
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));


function getPq(data) {
    var pq = []
    for (var i = 0; i < data.length; i++) {
        for (var j = 0; j < data.length; j++) {
            if (i === j) continue
            var keyStr = [
                data[i][0] - data[j][0],
                data[i][1] - data[j][1]
            ].join(',')
            pq.push(keyStr)
            // console.log(keyStr)
        }
    }
    pq.sort()
    var pqcount = {}
    for (var i = 0; i < pq.length; i++) {
        var key = pq[i]
        if (pqcount[key] == null) {
            pqcount[key] = 0
        }
        pqcount[key] += 1
    }
    var result = ""
    var resultValue = 0
    for (var key in pqcount) {
        if (resultValue < pqcount[key]) {
            resultValue = pqcount[key]
            result = key
        }
    }
    // console.log(data.length - resultValue)
    return result.split(",").map(Number)
}