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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<limits.h>

#define rep(i,begin,end) for(int i=begin; i<end; i++)
#define revrep(i,begin,end) for(int i=begin; i>end; i--)
#define lld long long int

typedef struct node{
    lld x, y;
}node;

int cmpAsc(const void* a, const void* b){
    lld x = ((node*)a)->y, y = ((node*)b)->y;
    if(x > y){
        return 1;
    }
    if(x < y){
        return -1;
    }
    return 0;
}

lld max(lld a, lld b, lld c, lld d){
    if(a >= b && a >= c && a >= d){
        return a;
    }
    if(b >= a && b >= c && b >= d){
        return b;
    }
    if(c >= a && c >= b && c >= d){
        return c;
    }
    return d;
}
lld min(lld a, lld b, lld c, lld d){
    if(a <= b && a <= c && a <= d){
        return a;
    }
    if(b <= a && b <= c && b <= d){
        return b;
    }
    if(c <= a && c <= b && c <= d){
        return c;
    }
    return d;
}

int main(){
    int n, k;
    scanf("%d %d", &n, &k);
    node a[n];
    rep(i, 0, n){
        lld x, y;
        scanf("%lld %lld", &x, &y);
        a[i].x = x;
        a[i].y = y;
    }
    lld ans = LLONG_MAX;
    lld top, bottom, left, right;
    rep(i, 0, n){
        rep(j, 0, n){
            rep(p, 0, n){
                rep(q, 0, n){
                    top    = max(a[i].y, a[j].y, a[p].y, a[q].y);
                    bottom = min(a[i].y, a[j].y, a[p].y, a[q].y);
                    left   = min(a[i].x, a[j].x, a[p].x, a[q].x);
                    right  = max(a[i].x, a[j].x, a[p].x, a[q].x);
                    int cnt = 0;
                    rep(r, 0, n){
                        if(bottom <= a[r].y && a[r].y <= top
                        && left   <= a[r].x && a[r].x <= right){
                            cnt++;
                        }
                    }
                    if(cnt >= k){
                        lld buf = (top - bottom) * (right - left);
                        if(buf < ans){
                            ans = buf;
                        }
                    }
                }
            }
        }
    }
    printf("%lld", ans);

    return 0;
}