Problem: 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}.

Examples

Input

2
0 3 D
3 0 L


Output

0


Input

5
-7 -10 U
7 -6 U
-8 7 D
-3 3 D
0 -6 R


Output

97.5


Input

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


Output

273

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

double inf = 1e10;

struct Data{
    vector<double> L = {inf, inf, inf};
    vector<double> R = {-inf, -inf, -inf};
    void add(int v, double w){
        L[v] = min(L[v], w);
        R[v] = max(R[v], w);
    }
    double get(double t){
        double nl = inf;
        double nr = -inf;
        for (int i = 0; i < 3; i++){
            nl = min(nl, L[i] + t * i);
            nr = max(nr, R[i] + t * i);
        }
        return nr - nl;
    }
    void event(vector<double> &times){
        for (int i = 0; i < 3; i++){
            for (int j = 0; j < i; j++){
                double t;
                t = (L[j] - L[i]) / (i - j);
                if (t > 0) times.push_back(t);
                t = (R[j] - R[i]) / (i - j);
                if (t > 0) times.push_back(t);
            }
        }
    }
};

int main(){
    int n;
    cin >> n;
    Data dx;
    Data dy;
    for (int i = 0; i < n; i++){
        int x, y; char d;
        cin >> x >> y >> d;
        if (d == 'R') dx.add(2, x), dy.add(1, y);
        if (d == 'L') dx.add(0, x), dy.add(1, y);
        if (d == 'U') dx.add(1, x), dy.add(2, y);
        if (d == 'D') dx.add(1, x), dy.add(0, y);
    }
    vector<double> times;
    times.push_back(0);
    times.push_back(inf);
    dx.event(times);
    dy.event(times);
    double ans = 1e18;
    for (int i = 0; i < times.size(); i++){
        ans = min(ans, dx.get(times[i]) * dy.get(times[i]));
    }
    cout << setprecision(20) << ans << endl;
}