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 _ ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long
#define pb push_back
#define sz(i) (int)(i.size())
#define x first
#define y second
#define L long double
#define P pair<int, int>
const int inf = 0x3f3f3f3f3f3f3f3LL;
const int mod = (int)1e9 + 7;
using namespace std;
#if LOCAL
#include "../../tr.h"
#else
#define tr(...)
#define endl '\n'
#endif
template<class T> void smin(T& a, T val) {if (a > val) a = val;}template<class T> void smax(T& a, T val) {if (a < val) a = val;}
const int N = (int)1e5 + 10;

#define double long double
int n;
char d[N];
P pt[N];

double f(double till){
   double x[2], y[2];
   x[0] = 1e20;//min
   x[1] = -1e20;//max
   y[0] = 1e20;
   y[1] = -1e20;

   for (int i = 0; i < n; ++i)
   {
      double ptx = pt[i].x, pty = pt[i].y;
      if (d[i] == 'R'){
         ptx += till;
      }
      else if (d[i] == 'L'){
         ptx -= till;
      }
      else if (d[i] == 'U'){
         pty += till;
      }
      else{
         pty -= till;
      }
      smax(x[1], ptx);
      smin(x[0], ptx);
      smax(y[1], pty);
      smin(y[0], pty);
   }

   return (x[1] - x[0]) * (y[1] - y[0]);
}

double binary_search(double A, double B){
   double ans = 0;

   for (int i = 0; i < 300; ++i)
   {
      double m1 = (A * 2 + B) / 3.0;
      double m2 = (A + 2 * B) / 3.0;
      if (f(m1) > f(m2)){
         ans = m2;
         A = m1;
      }
      else{
         B = m2;
      }
   }

   return f(ans);
}

int32_t main(){_

   cout << fixed << setprecision(15);
   cin>>n;

   for (int i = 0; i < n; ++i)
   {
      cin>>pt[i].x>>pt[i].y>>d[i];
   }   

   cout<<binary_search(0, 1e18);

   return 0;
}