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<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
typedef pair<int, int> P;

int n, k;
pair<int, int> p[50];

int main() {
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; i++) scanf("%d %d", &p[i].first, &p[i].second);
    sort(p, p + n);
    long ans = 4e18;
    for (int i = 0; i <= n - k; i++) {
        vector<int> y;
        for (int j = i; j < i + k - 1; j++) y.push_back(p[j].second);
        for (int j = i + k - 1; j < n; j++) {
            y.push_back(p[j].second);
            sort(y.begin(), y.end());
            int my = 2e9;
            for (int l = 0; l <= j - i + 1 - k; l++) {
                my = min(my, y[l + k - 1] - y[l]);
            }
            ans = min(ans, static_cast<long>(p[j].first - p[i].first) * my);
        }
    }

    printf("%ld\n", ans);
    return 0;
}