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;
double cxu,cxl,cyu,cyl;
bool bx,by;
double uR,lR,uL,lL,uU,lU,uD,lD;
bool bR,bL,bU,bD;
typedef struct{double x,y; char d;}dat;
auto u = [](double& a,double b){a=max(a,b);};
auto l = [](double& a,double b){a=min(a,b);};
const double inf = 1e10;
int main(){
  dat D;
  int n;
  cin>>n;
  for(int i=0;i<n;++i){
    cin>>D.x>>D.y>>D.d;
    switch(D.d){
      case 'U':
      case 'D':
        if(bx)u(cxu,D.x),l(cxl,D.x);
        else cxu=cxl=D.x,bx=1;
        break;
      case 'L':
      case 'R':
        if(by)u(cyu,D.y),l(cyl,D.y);
        else cyu=cyl=D.y,by=1;
    }
    switch(D.d){
      case 'U':
        if(bU)u(uU,D.y),l(lU,D.y);
        else uU=lU=D.y,bU=1;
        break;
      case 'D':
        if(bD)u(uD,D.y),l(lD,D.y);
        else uD=lD=D.y,bD=1;
        break;
      case 'L':
        if(bL)u(uL,D.x),l(lL,D.x);
        else uL=lL=D.x,bL=1;
        break;
      case 'R':
        if(bR)u(uR,D.x),l(lR,D.x);
        else uR=lR=D.x,bR=1;
        break;
    }
  }
  auto xmax = [&](double t){return max({bL?uL-t:-inf,bR?uR+t:-inf,bx?cxu:-inf});};
  auto xmin = [&](double t){return min({bL?lL-t:inf,bR?lR+t:inf,bx?cxl:inf});};
  auto ymax = [&](double t){return max({bD?uD-t:-inf,bU?uU+t:-inf,by?cyu:-inf});};
  auto ymin = [&](double t){return min({bD?lD-t:inf,bU?lU+t:inf,by?cyl:inf});};
  auto f = [&](double t){u(t,-t);return (xmax(t)-xmin(t))*(ymax(t)-ymin(t));};
  printf("%.12f\n",min({f(lL-cxl),f((lL-lR)/2.0),f(cxl-lR),f(uL-cxu),f((uL-uR)/2.0),f(cxu-uR),f(lD-cyl),f((lD-lU)/2.0),f(cyl-lU),f(uD-cyu),f((uD-uU)/2.0),f(cyu-uU),f(0.0)}));
}