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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
 
using namespace std;
 
#define INF 1000000007
#define LINF 1000000000000000007
 
typedef long long i64;
typedef pair<i64,i64> P;


int n;
double x,y;
char d;
double ma_u=-INF,mi_u=INF,ma_d=-INF,mi_d=INF,ma_l=-INF,mi_l=INF,ma_r=-INF,mi_r=INF;

int main(){
	cin >> n;
	double mi_x = INF, ma_x = 0, mi_y = INF, ma_y = 0;
	for(int i = 0; i < n; i++){
		cin >> x >> y >> d;
		if(d == 'U' || d == 'D'){
			mi_x = min(mi_x, x);
			ma_x = max(ma_x, x);
		}
		else{
			mi_y = min(mi_y, y);
			ma_y = max(ma_y, y);
		}
		
		if(d == 'U'){mi_u = min(mi_u,y); ma_u = max(ma_u,y);}
		if(d == 'D'){mi_d = min(mi_d,y); ma_d = max(ma_d,y);}
		if(d == 'L'){mi_l = min(mi_l,x); ma_l = max(ma_l,x);}
		if(d == 'R'){mi_r = min(mi_r,x); ma_r = max(ma_r,x);}
	}
	
	double ans = LINF;
	double cnt[4] = {0};
	if(mi_u <= mi_d){
		if(mi_d == INF) cnt[0] = max(0.0, ma_y - mi_u);
		cnt[0] = (mi_d - mi_u) / 2.0;
	}
	else{
		cnt[0] = max(0.0, mi_d - mi_y);
	}
	if(ma_u <= ma_d){
		if(ma_u == -INF) cnt[1] == max(0.0, ma_d - mi_y);
		cnt[1] = (ma_d - ma_u) / 2.0;
	}
	else{
		cnt[1] = max(0.0, ma_y - ma_u);
	}
	if(mi_r <= mi_l){
		if(mi_l == INF) cnt[2] == max(0.0, ma_x - mi_r);
		cnt[2] = (mi_l - mi_r) / 2.0;
	}
	else{
		cnt[2] = max(0.0, mi_l - mi_x);
	}
	if(ma_r <= ma_l){
		if(ma_r == -INF) cnt[3] == max(0.0, ma_l - mi_x);
		cnt[3] = (ma_l - ma_r) / 2.0;
	}
	else{
		cnt[3] = max(0.0, ma_x - ma_r);
	}
	for(int i = 0; i < 4; i++){
		ans = min(ans, (max(ma_y, max(ma_u+cnt[i], ma_d-cnt[i])) - min(mi_y, min(mi_d-cnt[i], mi_u+cnt[i]))) * (max(ma_x, max(ma_r+cnt[i], ma_l-cnt[i])) - min(mi_x, min(mi_l-cnt[i], mi_r+cnt[i]))));
	}
	printf("%.10f\n", ans);
	
	return 0;
}