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:
/// However long the night,
///     The dawn will break
/// ICPC next year

#include<bits/stdc++.h>
using namespace std ;
const int N = 1e5+5 ;
const long double eps = 1e-11;

int n ,x[N] ,y[N] ,addx[N] ,addy[N] ;
char ch ;
int xx[128] ,yy[128] ;
long double f(long double add){
    long double minX = 1e18 ,maxX = -1e18 ,minY = 1e18 ,maxY = -1e18 ;
    for(int i=0;i<n;++i){
        minX = min(minX ,(long double)x[i] + add*addx[i]);
        maxX = max(maxX ,(long double)x[i] + add*addx[i]);
        minY = min(minY ,(long double)y[i] + add*addy[i]);
        maxY = max(maxY ,(long double)y[i] + add*addy[i]);
    }
    return (maxX - minX) * (maxY - minY) ;
}
long double ternarySearch(long double left,long double right){
    for(int i=0;i<=1000;++i){
        long double g = left + (right - left) / 3, h = left + 2 * (right - left) / 3;
        if(f(g) < f(h))    right = h;
        else    left = g; // change < to > if the fn inc then dec
    }
    return f((left + right) / 2);
}
int main(){
    xx['R'] = 1 ;
    xx['L'] = -1 ;
    yy['U'] = 1 ;
    yy['D'] = -1 ;
    scanf("%d",&n);
    for(int i=0;i<n;++i){
        scanf("%d%d %c",x+i,y+i,&ch);
        addx[i] = xx[ch] ;
        addy[i] = yy[ch] ;
    }
    cout << fixed << setprecision(15) << ternarySearch(0,1e9+5);
    return 0;
}