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;
typedef long long ll;
 

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);
    
    
    ll n;
    cin>>n;
    double x[n],y[n];
    char d[n];
    for(int i=0;i<n;i++){
        cin>>x[i]>>y[i]>>d[i];
    }
    double dx[4] = {1,0,-1,0};
    double dy[4] = {0,1,0,-1};
    string s="RULD";    
    double INF = 1e18;
    double low = 0.0, up = 1e12;
    for(int c=0;c<1000;c++){
        double m[2] = {(low*2+up)/3,(low+up*2)/3};
        double x1[2]={INF,INF},x2[2]={-INF,-INF},y1[2]={INF,INF},y2[2]={-INF,-INF};
        for(int i=0;i<n;i++){
            for(int j=0;j<4;j++){
                if(d[i]==s[j]){
                    for(int k=0;k<2;k++){
                        x1[k] = min(x1[k], x[i]+dx[j]*m[k]);
                        x2[k] = max(x2[k], x[i]+dx[j]*m[k]);
                        y1[k] = min(y1[k], y[i]+dy[j]*m[k]);
                        y2[k] = max(y2[k], y[i]+dy[j]*m[k]);
                    }
                }
            }
        }
        double t1 = (x2[0] - x1[0]) * (y2[0] - y1[0]);
        double t2 = (x2[1] - x1[1]) * (y2[1] - y1[1]);
        if(t1 < t2){
            up = m[1];
        }
        else low = m[0];
    }
    double x1=INF, x2=-INF, y1=INF, y2=-INF;
    for(int i=0;i<n;i++){
        for(int j=0;j<4;j++){
            if(d[i]==s[j]){
                x1 = min(x1, x[i]+dx[j]*low);
                x2 = max(x2, x[i]+dx[j]*low);
                y1 = min(y1, y[i]+dy[j]*low);
                y2 = max(y2, y[i]+dy[j]*low);
            }
        }
    }
    double ans = (x2 - x1) * (y2 - y1);
    cout << ans << endl;
}