Problem:
Score : 600 points

Problem Statement

There are N points in a two-dimensional plane. The initial coordinates of the i-th point are (x_i, y_i). Now, each point starts moving at a speed of 1 per second, in a direction parallel to the x- or y- axis. You are given a character d_i that represents the specific direction in which the i-th point moves, as follows:

If d_i = R, the i-th point moves in the positive x direction;

If d_i = L, the i-th point moves in the negative x direction;

If d_i = U, the i-th point moves in the positive y direction;

If d_i = D, the i-th point moves in the negative y direction.

You can stop all the points at some moment of your choice after they start moving (including the moment they start moving).
Then, let x_{max} and x_{min} be the maximum and minimum among the x-coordinates of the N points, respectively. Similarly, let y_{max} and y_{min} be the maximum and minimum among the y-coordinates of the N points, respectively.

Find the minimum possible value of (x_{max} - x_{min}) \times (y_{max} - y_{min}) and print it.

Constraints

1 \leq N \leq 10^5

-10^8 \leq x_i,\ y_i \leq 10^8

x_i and y_i are integers.

d_i is R, L, U, or D.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1 d_1
x_2 y_2 d_2
.
.
.
x_N y_N d_N

Output

Print the minimum possible value of (x_{max} - x_{min}) \times (y_{max} - y_{min}).

The output will be considered correct when its absolute or relative error from the judge's output is at most 10^{-9}.

Sample Input 1

2
0 3 D
3 0 L

Sample Output 1

0

After three seconds, the two points will meet at the origin. The value in question will be 0 at that moment.

Sample Input 2

5
-7 -10 U
7 -6 U
-8 7 D
-3 3 D
0 -6 R

Sample Output 2

97.5

The answer may not be an integer.

Sample Input 3

20
6 -10 R
-4 -9 U
9 6 D
-3 -2 R
0 7 D
4 5 D
10 -10 U
-1 -8 U
10 -6 D
8 -5 U
6 4 D
0 3 D
7 9 R
9 -4 R
3 10 D
1 9 U
1 -6 U
9 -8 R
6 7 D
7 -3 D

Sample Output 3

273

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

void merge_sort(int x[], int n)
{
	static int y[100001] = {};
	if (n <= 1) return;
	merge_sort(&(x[0]), n/2);
	merge_sort(&(x[n/2]), (n+1)/2);
	
	int i, p, q;
	for (i = 0, p = 0, q = n/2; i < n; i++) {
		if (p >= n/2) y[i] = x[q++];
		else if (q >= n) y[i] = x[p++];
		else y[i] = (x[p] < x[q])? x[p++]: x[q++];
	}
	for (i = 0; i < n; i++) x[i] = y[i];
}

int main()
{
	const int sup = 2147483647, inf = -sup;
	int i, n[4] = {}, N, x, y, p[4][100001], max[2] = {inf, inf}, min[2] = {sup, sup};
	char d;
	scanf("%d", &N);
	for (i = 1; i <= N; i++) {
		scanf("%d %d %c", &x, &y, &d);
		if (d == 'R') {
			p[0][n[0]++] = x * 2;
			if (y * 2 > max[1]) max[1] = y * 2;
			if (y * 2 < min[1]) min[1] = y * 2;
		} else if (d == 'L') {
			p[1][n[1]++] = x * 2;
			if (y * 2 > max[1]) max[1] = y * 2;
			if (y * 2 < min[1]) min[1] = y * 2;
		} else if (d == 'U') {
			p[2][n[2]++] = y * 2;
			if (x * 2 > max[0]) max[0] = x * 2;
			if (x * 2 < min[0]) min[0] = x * 2;
		} else {
			p[3][n[3]++] = y * 2;
			if (x * 2 > max[0]) max[0] = x * 2;
			if (x * 2 < min[0]) min[0] = x * 2;
		}
	}
	merge_sort(p[0], n[0]);
	merge_sort(p[1], n[1]);
	merge_sort(p[2], n[2]);
	merge_sort(p[3], n[3]);
	
	int t[4][2];
	if (n[1] > 0 && p[1][n[1]-1] > max[0] && (n[0] == 0 || p[1][n[1]-1] > p[0][n[0]-1])) {
		if (max[0] == inf) t[0][0] = sup;
		else t[0][0] = p[1][n[1]-1] - max[0];
		if (n[0] > 0 && t[0][0] > (p[1][n[1]-1] - p[0][n[0]-1]) / 2) t[0][0] = (p[1][n[1]-1] - p[0][n[0]-1]) / 2;
	} else t[0][0] = 0;
	if (n[0] > 0) {
		if (max[0] <= p[0][n[0]-1]) t[0][1] = 0;
		else t[0][1] = max[0] - p[0][n[0]-1];
		if (n[1] > 0 && t[0][1] < (p[1][n[1]-1] - p[0][n[0]-1]) / 2) t[0][1] = (p[1][n[1]-1] - p[0][n[0]-1]) / 2;
	} else t[0][1] = sup;
	
	if (n[0] > 0 && p[0][0] < min[0] && (n[1] == 0 || p[0][0] < p[1][0])) {
		if (min[0] == sup) t[1][0] = sup;
		else t[1][0] = min[0] - p[0][0];
		if (n[1] > 0 && t[1][0] > (p[1][0] - p[0][0]) / 2) t[1][0] = (p[1][0] - p[0][0]) / 2;
	} else t[1][0] = 0;
	if (n[1] > 0) {
		if (min[0] >= p[1][0]) t[1][1] = 0;
		else t[1][1] = p[1][0] - min[0];
		if (n[0] > 0 && t[1][1] < (p[1][0] - p[0][0]) / 2) t[0][1] = (p[1][0] - p[0][0]) / 2;
	} else t[1][1] = sup;
	
	
	if (n[3] > 0 && p[3][n[3]-1] > max[1] && (n[2] == 0 || p[3][n[3]-1] > p[2][n[2]-1])) {
		if (max[1] == inf) t[2][0] = sup;
		else t[2][0] = p[3][n[3]-1] - max[1];
		if (n[2] > 0 && t[2][0] > (p[3][n[3]-1] - p[2][n[2]-1]) / 2) t[2][0] = (p[3][n[3]-1] - p[2][n[2]-1]) / 2;
	} else t[2][0] = 0;
	if (n[2] > 0) {
		if (max[1] <= p[2][n[2]-1]) t[2][1] = 0;
		else t[2][1] = max[2] - p[2][n[2]-1];
		if (n[3] > 0 && t[2][1] < (p[3][n[3]-1] - p[2][n[2]-1]) / 2) t[2][1] = (p[3][n[3]-1] - p[2][n[2]-1]) / 2;
	} else t[2][1] = sup;
	
	if (n[2] > 0 && p[2][0] < min[1] && (n[3] == 0 || p[2][0] < p[3][0])) {
		if (min[1] == sup) t[3][0] = sup;
		else t[3][0] = min[1] - p[2][0];
		if (n[3] > 0 && t[3][0] > (p[3][0] - p[2][0]) / 2) t[3][0] = (p[3][0] - p[2][0]) / 2;
	} else t[3][0] = 0;
	if (n[3] > 0) {
		if (min[1] >= p[3][0]) t[3][1] = 0;
		else t[3][1] = p[3][0] - min[1];
		if (n[2] > 0 && t[3][1] < (p[3][0] - p[2][0]) / 2) t[3][1] = (p[3][0] - p[2][0]) / 2;
	} else t[3][1] = sup;
	
	// for (i = 0; i < 4; i++) printf("%d %d\n", t[i][0], t[i][1]);
	
	int j;
	long long ans = (long long)sup * sup, tmp[4];
	for (i = 0; i < 4; i++) {
		for (j = 0; j < 2; j++) {
			tmp[0] = max[0];
			if (n[0] > 0 && tmp[0] < p[0][n[0]-1] + t[i][j]) tmp[0] = p[0][n[0]-1] + t[i][j];
			if (n[1] > 0 && tmp[0] < p[1][n[1]-1] - t[i][j]) tmp[0] = p[1][n[1]-1] - t[i][j];
			tmp[1] = min[0];
			if (n[0] > 0 && tmp[1] > p[0][0] + t[i][j]) tmp[1] = p[0][0] + t[i][j];
			if (n[1] > 0 && tmp[1] > p[1][0] - t[i][j]) tmp[1] = p[1][0] - t[i][j];
			tmp[2] = max[1];
			if (n[2] > 0 && tmp[2] < p[2][n[2]-1] + t[i][j]) tmp[2] = p[2][n[2]-1] + t[i][j];
			if (n[3] > 0 && tmp[2] < p[3][n[3]-1] - t[i][j]) tmp[2] = p[3][n[3]-1] - t[i][j];
			tmp[3] = min[1];
			if (n[2] > 0 && tmp[3] > p[2][0] + t[i][j]) tmp[3] = p[2][0] + t[i][j];
			if (n[3] > 0 && tmp[3] > p[3][0] - t[i][j]) tmp[3] = p[3][0] - t[i][j];
		
			if (ans > (tmp[0] - tmp[1]) * (tmp[2] - tmp[3])) ans = (tmp[0] - tmp[1]) * (tmp[2] - tmp[3]);
			// printf("%d %d %d %d %d %lld\n", t[i][j], tmp[0], tmp[1], tmp[2], tmp[3], ans);
		}
	}

	printf("%.1Lf\n", (long double)ans / 4.0);
	fflush(stdout);
	return 0;
}