Problem:
Score : 400 points

Problem Statement

We have N points in a two-dimensional plane.

The coordinates of the i-th point (1 \leq i \leq N) are (x_i,y_i).

Let us consider a rectangle whose sides are parallel to the coordinate axes that contains K or more of the N points in its interior.

Here, points on the sides of the rectangle are considered to be in the interior.

Find the minimum possible area of such a rectangle.

Constraints

2 \leq K \leq N \leq 50

-10^9 \leq x_i,y_i \leq 10^9 (1 \leq i \leq N)

x_i≠x_j (1 \leq i<j \leq N)

y_i≠y_j (1 \leq i<j \leq N)

All input values are integers. (Added at 21:50 JST)

Input

Input is given from Standard Input in the following format:

N K
x_1 y_1
:
x_{N} y_{N}

Output

Print the minimum possible area of a rectangle that satisfies the condition.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, K;
    cin >> N >> K;
    vector<pair<long long, long long>> xys(N);
    set<long long> xset;
    set<long long> yset;
    for (int i = 0; i < N; ++i){
        cin >> xys[i].first >> xys[i].second;
        xset.insert(xys[i].first);
    }
    vector<long long> xs(begin(xset), end(xset));
    long long ans = 9e18;
    for (auto xfirst = begin(xs); xfirst != end(xs); ++xfirst){
        for (auto xlast = xfirst; xlast != end(xs); ++xlast){
            vector<long long> candidates;
            for (auto xy : xys){
                long long x = xy.first;
                long long y = xy.second;
                if (x >= *xfirst and x <= *xlast) candidates.push_back(y);
            }
            long long x_span = *xlast - *xfirst;
            sort(begin(candidates), end(candidates));
            for (int head = 0; head + K <= candidates.size(); ++head){
                int tail = head + K - 1;
                ans = min(ans, (candidates[tail] - candidates[head]) * x_span);
            }
        }
    }
    cout << ans << endl;

    return 0;
}