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;
#define rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;
int main(){
    ll n,K,x,y;
    cin>>n>>K;
    vector<pair<ll,ll>> A(n);
    rep(i,n) cin>>x>>y,A[i]=make_pair(x,y);
    sort(A.begin(),A.end());
    ll m=4e18;
    rep(i,n)rep(j,n){
        if(j-i>=K-1){
            ll Xa=A[i].first,Xb=A[j].first;
            vector<ll> B;
            rep(k,j-i+1) B.push_back(A[i+k].second);
            sort(B.begin(),B.end());
            rep(k,j-i+2-K) m=min(m,(Xb-Xa)*(B[k+K-1]-B[k]));
        }
    }
    cout<<m;
}