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: Wrong Answer)

#include <bits/stdc++.h>

#include <boost/range/adaptors.hpp>
#include <boost/range/irange.hpp>

using namespace std;
using namespace boost;
using namespace boost::adaptors;

int main() {
    int64_t n, k;
    cin >> n >> k;

    vector<pair<int64_t, int64_t>> x(n), y(n);
    for (auto i : irange(0L, n)) {
        cin >> x[i].first >> y[i].first;
        x[i].second = y[i].second = i;
    }

    sort(x.begin(), x.end());
    sort(y.begin(), y.end());

    vector<pair<int64_t, int64_t>> points(n);
    for (auto i : irange(0L, n)) {
        points[x[i].second].first = i;
        points[y[i].second].second = i;
    }

    vector<vector<int64_t>> s(n + 1, vector<int64_t>(n + 1));
    for (auto i : irange(0L, n)) {
        ++s[points[i].first + 1][points[i].second + 1];
    }

    for (auto i : irange(0L, n + 1)) {
        for (auto j : irange(0L, n)) {
            s[i][j + 1] += s[i][j];
        }
    }

    for (auto i : irange(0L, n)) {
        for (auto j : irange(0L, n + 1)) {
            s[i + 1][j] += s[i][j];
        }
    }

    int64_t ans = numeric_limits<int64_t>::max();
    for (auto left : irange(0L, n)) {
        for (auto right : irange(left + 1, n)) {
            for (auto bottom : irange(0L, n)) {
                for (auto top : irange(bottom + 1, n)) {
                    auto num = s[top + 1][right + 1] - s[top + 1][left] -
                               s[bottom][right + 1] + s[bottom][left];

                    assert(num >= 0);

                    // cerr << left << "-" << right << "x" << bottom << "-" <<
                    // top
                    //      << ":" << num << endl;
                    if (num >= k) {
                        ans = min(ans,
                                  (x[right].first - x[left].first) *
                                          (y[top].first - y[bottom].first));
                    }
                }
            }
        }
    }

    assert(ans != numeric_limits<int64_t>::max());

    cout << ans << endl;
}