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 <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main(){
    int n,k;
    cin >> n >> k;
    vector<pair<long long, long long>> v;
    for (int i=0; i<n; i++){
        long long x,y;
        cin >> x >>y;
        v.push_back(make_pair(x,y));
    }
    sort(v.begin(),v.end());
    long long ans=10e19;
    for (int i=0; i<n-k+1; i++){
        for (int j=i+k-1; j<n; j++){
            vector<long long> vv;
            for (int x=i; x<=j; x++){
                vv.push_back(v[x].second);
            }
            sort(vv.begin(),vv.end());
            for (int l=0; l<j-i-k+2; l++){
                int m=l+k-1;
                long long scale=(v[j].first-v[i].first)*(vv[m]-vv[l]);
                ans=min(ans,scale);
            }
        }
    }
    cout << ans << endl;
}