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>
#define fastIO ios_base::sync_with_stdio(false),cin.tie(nullptr)
#define ll long long int
#define INF 5000000
#define eps 1e-13
#define mod 998244353
#define N 100005

using namespace std;

long double n,x[N],y[N],dx[N],dy[N];

long double calc(long double time)
{
    long double xMax=-1e18,xmin=1e18,yMax=-1e18,ymin=1e18,tmpx,tmpy;
    for(register int i=0;i<n;++i){
        tmpx=x[i]+time*dx[i];
        tmpy=y[i]+time*dy[i];

        xMax=max(xMax,tmpx);
        yMax=max(yMax,tmpy);

        xmin=min(xmin,tmpx);
        ymin=min(ymin,tmpy);
    }

    return (xMax-xmin)*(yMax-ymin);
}

inline void init()
{
    cin >> n;

    register char tmp;
    for(register int i=0;i<n;++i){
        cin >> x[i] >> y[i] >> tmp;

        switch(tmp){
            case'R':
                dx[i]=1;
                break;
            case'L':
                dx[i]=-1;
                break;
            case'U':
                dy[i]=1;
                break;
            case'D':
                dy[i]=-1;
                break;
        }
    }
}

inline void solve()
{
    long double TimeU=1e9,TimeD=0,ansMR,ansML,Ans=1e18;
    while(true){
        ansMR=calc( (TimeU*2+TimeD)/3 );
        ansML=calc( (TimeU+TimeD*2)/3 );

        Ans=min(ansML,Ans);
        Ans=min(ansMR,Ans);

        if(ansMR>ansML)
            TimeU=(TimeU*2+TimeD)/3;
        else
            TimeD=(TimeU+TimeD*2)/3;

        if(abs(ansMR-ansML)<eps)
            break;
    }

    cout << Ans << endl;
}

int main()
{
    fastIO;
    
    cout << fixed << setprecision(15);

    init();
    solve();

    cout << flush;

    return 0;
}