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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;
use std::collections::HashMap;
use std::cmp::min;

fn main() {
    input! {
        n:usize,
        xyd:[(i64,i64,char);n]
    };
    let mut dirmp = HashMap::new();
    dirmp.insert('R',0);
    dirmp.insert('L',1);
    dirmp.insert('U',2);
    dirmp.insert('D',3);

    let dxy:Vec<(i64,i64)> = vec![(1,0),(-1,0),(0,1),(0,-1)];

    let xyd:Vec<(i64,i64,usize)> = {
        xyd.into_iter().map(|(x,y,d)|(x*2,y*2,dirmp[&d])).collect()
    };

    let xyd = {
        let mut xyd_sml = Vec::new();
        for i in 0..4 {
            if let Some(e) = xyd.iter().filter(|&(_,_,d)|*d==i).max_by_key(|&(x,_,_)|x) {
                xyd_sml.push(*e);
            }
            if let Some(e) = xyd.iter().filter(|&(_,_,d)|*d==i).min_by_key(|&(x,_,_)|x) {
                xyd_sml.push(*e);
            }
            if let Some(e) = xyd.iter().filter(|&(_,_,d)|*d==i).max_by_key(|&(_,y,_)|y) {
                xyd_sml.push(*e);
            }
            if let Some(e) = xyd.iter().filter(|&(_,_,d)|*d==i).min_by_key(|&(_,y,_)|y) {
                xyd_sml.push(*e);
            }
        }
        xyd_sml
    };

    let ts = {
        let mut ts = vec![0];
        for &(x1,y1,d1) in &xyd {
            for &(x2,y2,d2) in &xyd {
                let dx = dxy[d1].0-dxy[d2].0;
                let dy = dxy[d1].1-dxy[d2].1;
                if dx!=0 {
                    let t = (x2-x1)/dx;
                    if 0<=t {
                        ts.push(t);
                    }
                }
                if dy!=0 {
                    let t = (y2-y1)/dy;
                    if 0<=t {
                        ts.push(t);
                    }
                }
            }
        }
        ts
    };

    let mut ans = std::i64::MAX;
    for t in ts {
        let max_x = xyd.iter().map(|&(x,_,d)|{
            let dx = dxy[d].0;
            x+dx*t
        }).max().unwrap();
        let min_x = xyd.iter().map(|&(x,_,d)|{
            let dx = dxy[d].0;
            x+dx*t
        }).min().unwrap();

        let max_y = xyd.iter().map(|&(_,y,d)|{
            let dy = dxy[d].1;
            y+dy*t
        }).max().unwrap();
        let min_y = xyd.iter().map(|&(_,y,d)|{
            let dy = dxy[d].1;
            y+dy*t
        }).min().unwrap();
        let sz = (max_x-min_x)*(max_y-min_y);
        ans = min(ans,sz);
    }

    println!("{}",ans as f64/4.0);

}