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;
using ll = long long;
const int inf = 1e9 + 7;
const ll longinf = 5LL << 60;
const ll mod = 1e9 + 7;

int main() {
    ll N, K;
    cin >> N >> K;
    vector<pair<ll, ll>> v(N);
    for(int i = 0; i < N; i++) cin >> v[i].first >> v[i].second;

    ll ans = longinf;
    for(int a = 0; a < N; a++) {
        for(int b = 0; b < N; b++) {
            for(int c = 0; c < N; c++) {
                for(int d = 0; d < N; d++) {
                    ll left = min(v[a].first, min(v[b].first, min(v[c].first, v[d].first)));
                    ll right = max(v[a].first, max(v[b].first, max(v[c].first, v[d].first)));
                    ll under = min(v[a].second, min(v[b].second, min(v[c].second, v[d].second)));
                    ll top = max(v[a].second, max(v[b].second, max(v[c].second, v[d].second)));

                    int cnt = 0;
                    for(int i = 0; i < N; i++) {
                        if(left <= v[i].first && v[i].first <= right && under <= v[i].second && v[i].second <= top) cnt++;
                    }
                    if(cnt >= K && (right - left) * (top - under) != 0) ans = min(ans, (right - left) * (top - under));
                }
            }
        }
    }
    cout << ans << "\n";
    return 0;
}