#include <bits/stdc++.h>
using namespace std;
vector<pair<pair<double,double>,int>> ps;
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

double f(double t){
    double mx_x=-9999999999;
    double mx_y=-9999999999;
    double mn_x=9999999999;
    double mn_y=9999999999;
    for(auto p:ps){
        mx_x=max(p.first.first+dx[p.second]*t,mx_x);
        mx_y=max(p.first.second+dy[p.second]*t,mx_y);
        mn_x=min(p.first.first+dx[p.second]*t,mn_x);
        mn_y=min(p.first.second+dy[p.second]*t,mn_y);
    }
    return -(mx_x-mn_x)*(mx_y-mn_y);
}

int main(void){
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        double x,y;
        string c;
        cin>>x>>y>>c;
        int v;
        if (c=="R") v=0;
        if (c=="U") v=1;
        if (c=="L") v=2;
        if (c=="D") v=3;
        ps.push_back(make_pair(make_pair(x,y),v));
    }
    double l=0.0,r=200000001.0;
    int nn=200;
    while(nn){
        if(f((l*2+r)/3.0)>f((l+r*2)/3.0))
            r=(l+r*2)/3.0;
        else
            l=(l*2+r)/3.0;
        nn--;
    }
    cout<<fixed<<setprecision(10)<<-f((l+r)/2)<<endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<pair<pair<double,double>,int>> ps;
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

double f(double t){
    double mx_x=-9999999999;
    double mx_y=-9999999999;
    double mn_x=9999999999;
    double mn_y=9999999999;
    for(auto p:ps){
        mx_x=max(p.first.first+dx[p.second]*t,mx_x);
        mx_y=max(p.first.second+dy[p.second]*t,mx_y);
        mn_x=min(p.first.first+dx[p.second]*t,mn_x);
        mn_y=min(p.first.second+dy[p.second]*t,mn_y);
    }
    return -(mx_x-mn_x)*(mx_y-mn_y);
}

int main(void){
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        double x,y;
        string c;
        cin>>x>>y>>c;
        int v;
        if (c=="R") v=0;
        if (c=="U") v=1;
        if (c=="L") v=2;
        if (c=="D") v=3;
        ps.push_back(make_pair(make_pair(x,y),v));
    }
    double l=0.0,r=200000001.0;
    int nn=200;
    while(nn){
        if(f((l*2+r)/3.0)>f((l+r*2)/3.0))
            r=(l+r*2)/3.0;
        else
            l=(l*2+r)/3.0;
        nn--;
    }
    cout<<fixed<<setprecision(10)<<-f((l+r)/2)<<endl;
}

```