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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define int long long
#define double long double
#pragma GCC optimize("O3,Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("avx,avx2,sse")
using namespace std;

int n;
vector<int> x(n);
vector<int> y(n);
vector<char> c(n);

double calc(double steps) {
    double ax = 1e15;
    double bx = -1e15;
    double ay = 1e15;
    double by = -1e15;
    for (int i = 0; i < n; i++) {
        double X = x[i];
        double Y = y[i];
        if (c[i] == 'L') X -= steps;
        if (c[i] == 'R') X += steps;
        if (c[i] == 'D') Y -= steps;
        if (c[i] == 'U') Y += steps;
        ax = min(ax, X);
        bx = max(bx, X);
        ay = min(ay, Y);
        by = max(by, Y);
    }
    return (by - ay) * (bx - ax);
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin >> n;
    x.resize(n);
    y.resize(n);
    c.resize(n);
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> y[i] >> c[i];
    }
    int s = 0;
    int e = 3e8;
    while (e - s > 5) {
        int a = (s * 2 + e) / 3;
        int b = (s + e * 2) / 3;
        double fa = calc(a);
        double fb = calc(b);
        if (fa <= fb) {
            e = b;
        } else {
            s = a;
        }
    }
    double res = LLONG_MAX;
    for (double i = s; i <= e; i += 0.5) {
        res = min(res, calc(i));
    }
    cout << res << '\n';

    fflush(stdout);
    return 0;
}