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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include <math.h>
#include <limits.h>
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) > (b) ? (b) : (a))
#define abs(x) ((x) > 0 ? (x) : -(x))
#define MOD (ll)1000000007 //10^9 + 7
#define endl printf("\n")
typedef long long ll;
#define MAX_N (1 << 17)

typedef struct{ll x, y;} point;

int
point_comp(const void *a, const void *b)
{
  point *A = (point *)a, *B = (point *)b;
  if (A->y > B->y) return 1;
  if (A->y < B->y) return -1;
  return 0;
}

int
main(int argc, char *argv[])
{
  int n, k;
  scanf("%d %d", &n, &k);
  point p[100];
  for (int i = 0; i < n; i++) scanf("%lld %lld", &p[i].x, &p[i].y);

  qsort(p, n, sizeof(point), point_comp);
  //for (int i = 0; i < n; i++) printf("(%lld, %lld)\n", p[i].x, p[i].y);

  ll ans = LLONG_MAX;
  for (int i = k - 1; i < n; i++) {
    ll xmin = LLONG_MAX, xmax = LLONG_MIN;
    for (int j = i - k + 1; j <= i; j++) {
      xmin = Min(xmin, p[j].x);
      xmax = Max(xmax, p[j].x);
    }
    ll tmp = (xmax - xmin) * (p[i].y - p[i - k + 1].y);
    ans = Min(ans, tmp);
  }

  printf("%lld\n", ans);

  return 0;
}