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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

int descending_compare(const void *a, const void *b){
    if (*(int*)a > *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

int ascending_compare(const void *a, const void *b){
    if (*(int*)a < *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

// array pointer = *a, num of element = n, key key
int lower_bound(int *a, int n, int key){
    int ng, mid, ok;
    ng = -1, ok = n-1;
    while (abs(ok - ng) > 1){
        mid = (ok + ng) / 2;
        if (key <= a[mid]){
            ok = mid;
        }else{
            ng = mid;
        }
    }
    if (a[ok] >= key)return ok;
    return n;
}

//greatest common divisor
unsigned long gcd(unsigned long x, unsigned long y){
    if (y == 0){ 
        return x;
    }else if (x > y){
        return gcd(y, x % y);
    }else{
        return gcd(x, y % x);
    }
}

unsigned long lcm(unsigned long x, unsigned long y){
    unsigned long g = gcd(x, y);
    return x*y/g;

}



long long factorial(int x){
    long long rtn = 1;
    int i;
    for (i = x; i > 1; i--){
        rtn = (rtn*i);
    }
    return rtn;
}

/*
int struct_ascending_compare(const void *p, const void *q) {
return ((struct_name*)p)->member_name - ((struct_name*)q)->member_name;
}*/

/*unsigned long long pascal[100][100] = {0};
void make_pascal(void){
    for (int i = 0; i < 100; i++){
        pascal[i][0] = 1;
    }
    pascal[1][1] = 1;
    for (int i = 2; i < 100; i++){
        for (int j = 1; j < 100; j++){
           pascal[i][j] = (pascal[i-1][j-1]+pascal[i-1][j]) % mod;
        }
    }
}*/
long long mod = 1000000007;

//x ^ n
long long mod_pow(long long x, long long n){
    long long res = 1;
    for(int i = 0;i < 60; i++){
        if(n >> i & 1) res = res * x % mod;
        x = x * x % mod;
    }
    return res;
}


int x[55];
int y[55];

int main(void){
    int n, k;
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; i++){
        scanf("%d %d", &x[i], &y[i]);
    }
    long long ans = 3999999999000000001;
    for (int i = 0; i < n; i++){
        for (int j = i + 1; j < n; j++){
            int hx = x[i]>x[j]?x[i]:x[j];
            int lx = x[i]<x[j]?x[i]:x[j];
            int hy = y[i]>y[j]?y[i]:y[j];
            int ly = y[i]<y[j]?y[i]:y[j];
            int cnt = 0;
            for (int k = 0; k < n; k++){
                if (lx <= x[k] && hx >= x[k] && ly <= y[k] && hy >= y[k]) cnt++;
            }
            if (cnt >= k) ans = ans>llabs(hx-lx)*llabs(hy-ly)?llabs(hx-lx)*llabs(hy-ly):ans;
        }
    }

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

}