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 (1.15.1) (judge verdict: Accepted)

//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

//

fn run() {
    input! {
        n: usize,
        p: [(f64, f64, chars); n],
    }
    let mut map = std::collections::BTreeMap::new();
    map.insert('R', 0);
    map.insert('U', 1);
    map.insert('L', 2);
    map.insert('D', 3);
    let mut p: Vec<_> = p.into_iter().map(|p| (p.0, p.1, *map.get(&p.2[0]).unwrap())).collect();
    let mut a = vec![];
    a.push(0f64);
    for _ in 0..4 {
        let inf = -100_000_000f64;
        let mut left = inf;
        let mut fix = inf;
        let mut right = inf;
        for &(x, _, k) in p.iter() {
            if k == 0 {
                right = right.max(x);
            } else if k == 2 {
                left = left.max(x);
            } else {
                fix = fix.max(x);
            }
        }
        for &v in [left - fix, fix - right, (left - right) * 0.5].iter() {
            if v > 0.0 {
                a.push(v);
            }
        }
        for p in p.iter_mut() {
            *p = (-p.1, p.0, (p.2 + 1) % 4);
        }
    }
    let mut ans = std::f64::MAX;
    let dir = [(1.0, 0.0), (0.0, 1.0), (-1.0, 0.0), (0.0, -1.0)];
    for a in a {
        let mut d = std::f64::MAX;
        let mut u = std::f64::MIN;
        let mut l = std::f64::MAX;
        let mut r = std::f64::MIN;
        for &(x, y, k) in p.iter() {
            d = d.min(y + a * dir[k].1);
            u = u.max(y + a * dir[k].1);
            l = l.min(x + a * dir[k].0);
            r = r.max(x + a * dir[k].0);
        }
        ans = ans.min((u - d) * (r - l));
    }
    println!("{:.10}", ans);
}

fn main() {
    run();
}