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 <cstdio>
#include <algorithm>
#include <vector>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)

using ll = long long;

ll N, K;
std::vector<ll> x, y;

int main()
{
    scanf( "%lld%lld", &N, &K );
    rep( i, N )
    {
      ll tx, ty;
      scanf( "%lld%lld", &tx, &ty );
      x.push_back( tx );
      y.push_back( ty );
    }

    ll ans = 5000000000000000000;

    rep( i, N ) rep( j, N ) rep( k, N ) rep( l, N )
    {
        //[i,j] [k,l]

        ll lx = std::min( x[i], x[j] ), rx = std::max( x[i], x[j] );
        ll ly = std::min( y[k], y[l] ), ry = std::max( y[k], y[l] );

        int cnt = 0;

        rep( i2, N ) if( lx <= x[i2] && x[i2] <= rx && ly <= y[i2] && y[i2] <= ry )
          ++cnt;

        if( cnt >= K )
            ans = std::min( ans, 1ll*(rx-lx)*(ry-ly) );
    }

    printf( "%lld\n", ans );

    return 0;
}