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: Wrong Answer)

#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <set>
#include <queue>
#include <map>
#include <numeric>
#include <functional>
#include <math.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;

#define INF ((int)1e9)
#define INFLL ((ll)1e18)
#define MOD (1000000007LL)

enum Dir {
    R, L, U, D,
};

int main(int argc, const char * argv[]) {
    ios_base::sync_with_stdio(false);
    ll MinX[4] = {INFLL, INFLL, INFLL, INFLL};
    ll MaxX[4] = {-INFLL, -INFLL, -INFLL, -INFLL};
    ll MinY[4] = {INFLL, INFLL, INFLL, INFLL};
    ll MaxY[4] = {-INFLL, -INFLL, -INFLL, -INFLL};
    int N;
    cin >> N;
    for (int n = 0; n < N; ++n) {
        int x, y;
        char cd;
        Dir d;
        cin >> x >> y >> cd;
        switch (cd) {
            case 'R': d = R; break;
            case 'L': d = L; break;
            case 'U': d = U; break;
            case 'D': d = D; break;
            default: break;
        }
        if (MinX[d] > x) MinX[d] = x;
        if (MaxX[d] < x) MaxX[d] = x;
        if (MinY[d] > y) MinY[d] = y;
        if (MaxY[d] < y) MaxY[d] = y;
    }
    
    auto Calc = [&](double T)
    {
        double X, Y;
        {
            double Min = INFLL;
            Min = min(Min, MinX[R]+T);
            Min = min(Min, MinX[L]-T);
            Min = min(Min, MinX[U]+0.0);
            Min = min(Min, MinX[D]+0.0);
            double Max = -INFLL;
            Max = max(Max, MaxX[R]+T);
            Max = max(Max, MaxX[L]-T);
            Max = max(Max, MaxX[U]+0.0);
            Max = max(Max, MaxX[D]+0.0);
            X = Max - Min;
        }
        {
            double Min = INFLL;
            Min = min(Min, MinY[R]+0.0);
            Min = min(Min, MinY[L]+0.0);
            Min = min(Min, MinY[U]+T);
            Min = min(Min, MinY[D]-T);
            double Max = -INFLL;
            Max = max(Max, MaxY[R]+0.0);
            Max = max(Max, MaxY[L]+0.0);
            Max = max(Max, MaxY[U]+T);
            Max = max(Max, MaxY[D]-T);
            Y = Max - Min;
        }
        return X * Y;
    };
    
    double l = 0.0, r = 1e10;
    while (r - l > 1e-13 && r - l > r * 1e-13) {
        double x1 = (2*l+r) / 3;
        double x2 = (l+2*r) / 3;
        if (Calc(x1) < Calc(x2)) {
            r = x2;
        } else {
            l = x1;
        }
    }
    cout << fixed;
    cout.precision(15);
    cerr << l << endl;
    cout << Calc((l+r)/2) << endl;
    
    return 0;
}