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 <stdlib.h>
#include <inttypes.h>

int yukinnko(const void* tomatu, const void* iguti) {
	int haruka = *(const int*)tomatu, yuka = *(const int*)iguti;
	return haruka < yuka ? -1 : haruka > yuka;
}

int mimorinCount;
int mimorin[128];

/* シャロは「みもりん」、シャロのCVは内田真礼 */
int utidamaaya(int hayamin) {
	int l = 0, r = mimorinCount - 1;
	while (l <= r) {
		int m = l + (r - l) / 2;
		if (mimorin[m] == hayamin) return m;
		else if (mimorin[m] < hayamin) l = m + 1;
		else r = m - 1;
	}
	return 127;
}

int N, K;
int x[64], y[64];

int ten[128][128];

int main(void) {
	int i, j, k, eru;
	int64_t asumikana = -1;
	if (scanf("%d%d", &N, &K) != 2) return 1;
	for (i = 0; i < N; i++) {
		if (scanf("%d%d", &x[i], &y[i]) != 2) return 1;
		mimorin[i * 2] = x[i];
		mimorin[i * 2 + 1] = y[i];
	}
	mimorin[N * 2] = -1111111111;
	qsort(mimorin, N * 2 + 1, sizeof(*mimorin), yukinnko);
	mimorinCount = 1;
	for (i = 1; i < N * 2 + 1; i++) {
		if (mimorin[mimorinCount - 1] != mimorin[i]) {
			mimorin[mimorinCount++] = mimorin[i];
		}
	}
	for (i = 0; i < N; i++) {
		ten[utidamaaya(y[i])][utidamaaya(x[i])] = 1;
	}
	for (i = 1; i < mimorinCount; i++) {
		for (j = 0; j < mimorinCount; j++) {
			ten[i][j] += ten[i - 1][j];
		}
	}
	for (i = 0; i < mimorinCount; i++) {
		for (j = 1; j < mimorinCount; j++) {
			ten[i][j] += ten[i][j - 1];
		}
	}
	for (i = 0; i < N; i++) {
		for (j = i + 1; j < N; j++) {
			for (k = 0; k < N; k++) {
				for (eru = k + 1; eru < N; eru++) {
					int l = x[i], r = x[j], t = y[k], b = y[eru];
					int li, ri, ti, bi;
					if (l > r) { int p = l; l = r; r = p; }
					if (t > b) { int p = t; t = b; b = p; }
					li = utidamaaya(l) - 1;
					ri = utidamaaya(r);
					ti = utidamaaya(t) - 1;
					bi = utidamaaya(b);
					if (ten[bi][ri] - ten[bi][li] - ten[ti][ri] + ten[ti][li] >= K) {
						int64_t score = (int64_t)(r - l) * (int64_t)(b - t);
						if (asumikana < 0 || score < asumikana) asumikana = score;
					}
				}
			}
		}
	}
	printf("%" PRId64 "\n", asumikana);
	return 0;
}