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>
using namespace std;

#define DUMP(x) cerr << #x << "=" << x << endl
#define DUMP2(x, y) cerr<<"("<<#x<<", "<<#y<<") = ("<<x<<", "<<y<<")"<< endl
#define BINARY(x) static_cast<bitset<16> >(x)

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,m,n) for (int i=m;i<(int)(n);i++)

#define in_range(x, y, w, h) (0<=(int)(x) && (int)(x)<(int)(w) && 0<=(int)(y) && (int)(y)<(int)(h))
#define ALL(a) (a).begin(),(a).end()

typedef long long ll;
const int INF   = 1e9;
const ll  INFLL = 5e18;
typedef pair<ll,ll> PII;
typedef vector<vector<int>> Graph;
int dx[4]={0, -1, 1, 0}, dy[4]={-1, 0, 0, 1};

int main()
{
    ios::sync_with_stdio(false);

    ll N, K;
    cin >> N >> K;
    vector<PII> P(N);
    rep(i, N) {
        cin >> P[i].first >> P[i].second;
    }

    sort(ALL(P));

    ll ans = INFLL;
    for (int i=0; i<=N-K; i++) {
        ll min_x = INFLL, max_x = -INFLL, min_y = INFLL, max_y = -INFLL;
        rep(j, K) {
            min_x = min(min_x, P[i + j].first);
            max_x = max(max_x, P[i + j].first);
            min_y = min(min_y, P[i + j].second);
            max_y = max(max_y, P[i + j].second);
        }
        ans = min(ans, (max_x - min_x) * (max_y - min_y));
    }
    cout << ans << endl;
}