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: Time Limit Exceeded)

#include <stdio.h>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <list>
#include <iterator>
#include <assert.h>
#pragma warning(disable:4996) 

typedef long long ll;
#define MIN(a, b) ((a)>(b)? (b): (a))
#define MAX(a, b) ((a)<(b)? (b): (a))
#define LINF 9223300000000000000
#define INF 2140000000
const long long MOD = 1000000007;
using namespace std;

int a[2][3][2];    // (x/y)(-/0/+)(min/max)



int main(int argc, char* argv[])
{
    int n;
    scanf("%d", &n);

    int i,j;
    for(i=0; i<2; i++) {
        for(j=0; j<3; j++) {
            a[i][j][0]=INF;  a[i][j][1]=-INF;
        }
    }

    for(i=0; i<n; i++) {
        int x,y;
        char str[3]={0};
        scanf("%d%d%s", &x, &y, str);
        if(str[0]=='R') {
            a[0][2][0]=MIN(a[0][2][0], x);
            a[0][2][1]=MAX(a[0][2][1], x);
            a[1][1][0]=MIN(a[1][1][0], y);
            a[1][1][1]=MAX(a[1][1][1], y);
        }
        else if(str[0]=='L') {
            a[0][0][0]=MIN(a[0][0][0], x);
            a[0][0][1]=MAX(a[0][0][1], x);
            a[1][1][0]=MIN(a[1][1][0], y);
            a[1][1][1]=MAX(a[1][1][1], y);
        }
        else if(str[0]=='U') {
            a[0][1][0]=MIN(a[0][1][0], x);
            a[0][1][1]=MAX(a[0][1][1], x);
            a[1][2][0]=MIN(a[1][2][0], y);
            a[1][2][1]=MAX(a[1][2][1], y);
        }
        else {
            a[0][1][0]=MIN(a[0][1][0], x);
            a[0][1][1]=MAX(a[0][1][1], x);
            a[1][0][0]=MIN(a[1][0][0], y);
            a[1][0][1]=MAX(a[1][0][1], y);
        }
    }

    double ans = (double)INF * (double)INF;
    int t,k;
    for(t=0; t<=400000000; t++) {
        double range[2][2];
        for(i=0; i<2; i++) {
            range[i][0]=INF; range[i][1]=-INF;
        }

        for(i=0; i<2; i++) {
            for(j=0; j<3; j++) {
                for(k=0; k<2; k++) {
                    double tmp = (double)a[i][j][k];
                    if(j==0) tmp -= (double)t/2;
                    else if(j==2) tmp += (double)t/2;

                    if(k==0) range[i][k]=MIN(range[i][k], tmp);
                    else     range[i][k]=MAX(range[i][k], tmp);
                }
            }
        }
        double val = (range[0][1]-range[0][0])*(range[1][1]-range[1][0]);
        ans = MIN(ans, val);
    }
    printf("%.10lf\n", ans);

    return 0;
}