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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int mod=1000000007,MAX=100001,INF=1<<30;

int main(){

    int N;cin>>N;
    vector<long double> X(N),Y(N);
    vector<char> S(N);
    for(int i=0;i<N;i++){
        cin>>X[i]>>Y[i]>>S[i];
    }
    long double UX1,LX1,RX1,DX1,UX2,LX2,RX2,DX2,UY1,LY1,RY1,DY1,UY2,LY2,RY2,DY2;//1は正
    UX1=LX1=RX1=DX1=UY1=LY1=RY1=DY1=-INF;
    UX2=LX2=RX2=DX2=UY2=LY2=RY2=DY2=INF;
    
    for(int i=0;i<N;i++){
        if(S[i]=='U'){
            UX1=max(UX1,X[i]);
            UX2=min(UX2,X[i]);
            UY1=max(UY1,Y[i]);
            UY2=min(UY2,Y[i]);
        }else if(S[i]=='L'){
            LX1=max(LX1,X[i]);
            LX2=min(LX2,X[i]);
            LY1=max(LY1,Y[i]);
            LY2=min(LY2,Y[i]);
        }else if(S[i]=='R'){
            RX1=max(RX1,X[i]);
            RX2=min(RX2,X[i]);
            RY1=max(RY1,Y[i]);
            RY2=min(RY2,Y[i]);
        }else{
            DX1=max(DX1,X[i]);
            DX2=min(DX2,X[i]);
            DY1=max(DY1,Y[i]);
            DY2=min(DY2,Y[i]);
        }
    }
    
    long double ans;
    
    long double xmax=max({UX1,LX1,RX1,DX1});
    long double xmin=min({UX2,LX2,RX1,DX2});
    long double ymax=max({UY1,LY1,RY1,DY1});
    long double ymin=min({UY2,LY2,RY2,DY2});
    
    long double a=(xmax-xmin)*(ymax-ymin);
    ans=a;
    
    for(long double t=0.5;t<=100000000.0;t+=0.5){
        long double xmax=max({UX1,LX1-t,RX1+t,DX1});
        long double xmin=min({UX2,LX2-t,RX1+t,DX2});
        long double ymax=max({UY1+t,LY1,RY1,DY1-t});
        long double ymin=min({UY2+t,LY2,RY2,DY2-t});
        
        long double a=(xmax-xmin)*(ymax-ymin);
        ans=min(ans,a);
    }
    
    cout<<setprecision(25)<<ans<<endl;
    
}