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: Wrong Answer)

// ---------- begin chmin, chmax ----------
trait ChangeMinMax {
    fn chmin(&mut self, x: Self) -> bool;
    fn chmax(&mut self, x: Self) -> bool;
}

impl<T: PartialOrd> ChangeMinMax for T {
    fn chmin(&mut self, x: Self) -> bool {
        *self > x && {
            *self = x;
            true
        }
    }
    fn chmax(&mut self, x: Self) -> bool {
        *self < x && {
            *self = x;
            true
        }
    }
}
// ---------- end chmin, chmax ----------

use proconio::marker::*;
use proconio::*;
use std::cmp::*;
use std::collections::*;

fn run() {
    input! {
        n: usize,
        p: [(f64, f64, Chars); n],
    }
    let map = vec![('U', 0), ('L', 1), ('D', 2), ('R', 3)]
        .into_iter()
        .collect::<BTreeMap<_, _>>();
    let dir = [(0f64, 1f64), (-1f64, 0f64), (0f64, -1f64), (1f64, 0f64)];
    let mut p = p
        .into_iter()
        .map(|(x, y, s)| (x, y, map.get(&s[0]).cloned().unwrap()))
        .collect::<Vec<_>>();
    let inf = 10f64.powi(9);
    let mut ans = inf * inf;
    for _ in 0..4 {
        let mut down = -inf;
        let mut up = -inf;
        let mut fix = -inf;
        for &(_, y, d) in p.iter() {
            if d == 0 {
                up.chmax(y);
            } else if d == 2 {
                down.chmax(y);
            } else {
                fix.chmax(y);
            }
        }
        for &t in [down - fix, (down - up) * 0.5, fix - up].iter() {
            if t < 0f64 {
                continue;
            }
            let mut range = (inf, -inf, inf, -inf);
            for &(x, y, d) in p.iter() {
                let x = x + dir[d].0 * t;
                let y = y + dir[d].1 * t;
                range.0.chmin(x);
                range.1.chmax(x);
                range.2.chmin(y);
                range.3.chmax(y);
            }
            ans.chmin((range.1 - range.0) * (range.3 - range.2));
        }
        for p in p.iter_mut() {
            *p = (p.1, -p.0, (p.2 + 1) & 3);
        }
    }
    println!("{:.2}", ans);
}

fn main() {
    run();
}