Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

# include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=100;
const LL inf=(4*1e9)*(4*1e9);
LL x[maxn], y[maxn];
int n, k;

int check(LL px1, LL py1, LL px2, LL py2){
    int cnt=0;
    for(int i=0;i<n;i++){
        bool bx=1, by=1;
        if((x[i]>px1&&x[i]>px2)||(x[i]<px1&&x[i]<px2))bx=0;
        if((y[i]>py1&&y[i]>py2)||(y[i]<py1&&y[i]<py2))by=0;
        if(bx&&by)cnt++;
    }
    return cnt;
}

LL pd(LL x1, LL x2){
    return max(x1, x2)-min(x1, x2);
}

int main(){
    while(~scanf("%d%d", &n, &k)){
        LL maxx=-inf,maxy=-inf,minx=inf,miny=inf;
        for(int i=0;i<n;i++){
            scanf("%lld%lld", &x[i], &y[i]);
            maxx=max(maxx, x[i]);
            maxy=max(maxy, y[i]);
            minx=min(minx, x[i]);
            miny=min(miny, y[i]);
        }
        LL ans=inf;
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                if(check(x[i], y[i], x[j], y[j])>=k){
                    ans=min(ans, pd(x[i],x[j])*pd(y[i], y[j]));
                }
                for(int a=j+1;a<n;a++){
                    LL upx=a[x], lowx=a[x], upy=a[y], lowy=a[y];
                    upx=max(upx, x[i]),upx=max(upx, x[j]);
                    lowx=min(lowx, x[i]),lowx=min(lowx,x[j]);
                    upy=max(upy, y[i]),upy=max(upy, y[j]);
                    lowy=min(lowx, x[i]),lowy=min(lowy,y[j]);
                    if(check(upx, upy, lowx, lowy)>=k){
                        ans=min(ans, pd(upx,lowx)*pd(upy, lowy));
                    }
                }
            }
        }
        if(ans==inf)ans=pd(maxx, minx)*pd(maxy, miny);
        printf("%lld\n", ans);
    }
    return 0;
}