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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdio>

using namespace std;

int main(){
    int N;
    while(cin >> N){
        vector<double> times;
        times.push_back(0);
        vector<long long> x(N), y(N);
        vector<string> d(N);
        for(int i=0;i<N;i++) cin >> x[i] >> y[i] >> d[i];
        vector<int> dx(N, 0), dy(N, 0);
        for(int i=0;i<N;i++){
            if(d[i] == "L") dx[i] = -1;
            if(d[i] == "R") dx[i] = 1;
            if(d[i] == "U") dy[i] = 1;
            if(d[i] == "D") dy[i] = -1;
        }
        auto calcArea = [&](double t){
            double minX = 1e20, maxX = -1e20, minY = 1e20, maxY = -1e20;
            for(int j=0;j<N;j++){
                double cx = x[j] + t * dx[j];
                double cy = y[j] + t * dy[j];
                minX = min(cx, minX);
                maxX = max(cx, maxX);
                minY = min(cy, minY);
                maxY = max(cy, maxY);
            }                
            return (maxX - minX) * (maxY - minY);
        };
        double res = calcArea(0);
            double l = 0, r = 1e9;
            for(int _=0;_<1000;_++){
                double midA = (2*l+r)/3;
                double midB = (l+2*r)/3;
                double areaA = calcArea(midA);
                double areaB = calcArea(midB);
                if(areaA < areaB) r = midB;
                else l = midA;
            }
            res = min(res, calcArea(0.5*(l+r)));
        // }
        printf("%.9lf\n", res);
    }
}