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 <iostream>
#include <vector>
#include <utility>
#include <algorithm>
#define ll long long
using namespace std;

int main() {
    int N, K; cin >> N >> K;
    vector<pair<ll, ll>> v;
    for(int i=0; i<N; ++i){
        ll x, y;
        cin >> x >> y;
        v.push_back(make_pair(x, y));
    }
    sort(v.begin(), v.end());
    ll ans=5e18;
    for(int i=0; i+K<=N; ++i){
        vector<ll> u;
        for(int j=0; j+1<K; ++j) u.push_back(v[i+j].second);
        for(int j=i+K-1; j<N; ++j){
            u.push_back(v[j].second);
            sort(u.begin(), u.end());
            ll mn=3e9;
            for(int t=0; t+K-1<=j-i; ++t) mn=min(mn, u[t+K-1]-u[t]);
            ans=min(ans, mn*(v[j].first-v[i].first));
        }
    }
    cout << ans << endl;
    return 0;
}