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;
using ll = long long;
template <class T> using vec = vector<T>;
template <class T> using vvec = vector<vec<T>>;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N,K;
    cin >> N >> K;
    vec<ll> X(N),Y(N);
    vec<ll> x(N),y(N);
    for(int i=0;i<N;i++){
        cin >> X[i] >> Y[i];
        x[i] = X[i],y[i] = Y[i];
    }
    ll ans = 4e18;
    sort(x.begin(),x.end());
    sort(y.begin(),y.end());
    for(int a=0;a<N;a++) for(int b=a+1;b<N;b++){
        for(int c=0;c<N;c++) for(int d=c+1;d<N;d++){
            int cnt = 0;
            for(int i=0;i<N;i++) cnt += x[a]<=X[i] && X[i]<=x[b] && y[c]<=Y[i] && Y[i]<=y[d];
            if(cnt>=K) ans = min(ans,(x[b]-x[a])*(y[d]-y[c]));
        }
    }
    cout << ans << "\n";
}