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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define zero_pad(num) setfill('0') << std::right << setw(num)
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;

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

    ll ans = 4001001001001001001;
    rep(i, n-1)for(int j = i+1; j < n; j++){
        if(j-i+1 < k)continue;
        vector<ll> yy;
        for(int l = i; l <= j; l++)yy.push_back(x[l].second);
        sort(yy.begin(), yy.end());
        ll dy = 2002002002;
        rep(l, yy.size()-k+1)dy = min(dy, yy[l+k-1]-yy[l]);
        ans = min(ans, dy * (x[j].first-x[i].first));
    }
    cout << ans << endl;
}