Problem: 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.

Examples

Input

4 4
1 4
3 3
6 2
8 1


Output

21


Input

4 2
0 0
1 1
2 2
3 3


Output

1


Input

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


Output

3999999996000000001

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef pair<int, int> P;

lint x[55], y[55];
vector<P> point;

int main(){
    int n, K; cin >> n >> K;
    for(int i=0;i<n;i++){
        cin >> x[i] >> y[i];
        point.push_back(P(x[i], y[i]));
    }

    sort(x, x+n); sort(y, y+n);

    lint ans = 1e20;
    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            for(int k=0;k<n;k++){
                for(int l=k+1;l<n;l++){
                    int cnt = 0;
                    for(auto it : point)
                        if(x[i]<=it.first && it.first<=x[j] && y[k]<=it.second && it.second<=y[l]) cnt++;
                    if(K<=cnt) ans = min(ans, (x[j]-x[i])*(y[l]-y[k]));
                }
            }
        }
    }

    cout << ans << endl;
    return 0;
}