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;
#define rep(i,n) for(int i=0;i<n;++i)
#define rep2(i,i0,n) for(int i=i0;i<n;++i)

int main()
{
    int n,k;
    cin >> n >> k;
    vector<pair<ll,ll>> xy(n);
    rep(i,n) {
        cin >> xy[i].first >> xy[i].second;
    }
    sort(xy.begin(), xy.end());

    ll ans = INT64_MAX;
    for(int i=0;i<n-k+1;++i) {
        for(int j=i+k-1;j<n;++j) {
            vector<ll> y;
            for(int l=i;l<=j;++l) 
                y.push_back(xy[l].second);
            sort(y.begin(),y.end());

            ll dy = INT64_MAX;
            int l_max = y.size()-k+1;
            for(int l=0;l<l_max;++l)
                dy = min(dy, y[l+k-1]-y[l]);

            ans = min(ans, (xy[j].first-xy[i].first)*dy);
        }
    }

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