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 <cassert>
#include <cstdio>
#include <iostream>
using namespace std;
#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
typedef long long ll;


double xs[100000+5];
double ys[100000+5];
char ds[100000+5];
int n;

double calc(double t) {
  double x1 = 1e10;
  double x2 = -1e10;
  double y1 = 1e10;
  double y2 = -1e10;
  REP(i, n) {
    double x, y;
    if(ds[i] == 'U') {
      x = xs[i];
      y = ys[i] + t;
    } else if(ds[i] == 'D') {
      x = xs[i];
      y = ys[i] - t;
    } else if(ds[i] == 'L') {
      x = xs[i] - t;
      y = ys[i];
    } else if(ds[i] == 'R') {
      x = xs[i] + t;
      y = ys[i];
    } else {
      assert(false);
    }
    x1 = min(x1, x);
    y1 = min(y1, y);
    x2 = max(x2, x);
    y2 = max(y2, y);
  }
  double res = (x2-x1)*(y2-y1);
  // cerr << t << " " << res << endl;
  return res;
}

int main(void) {
  scanf("%d", &n);
  REP(i, n) {
    scanf("%lf%lf %c", &xs[i], &ys[i], &ds[i]);
  }
  double left = 0;
  double right = 1e9;
  REP(t, 200) {
    double m1 = (left*2+right) / 3.0;
    double m2 = (left+right*2) / 3.0;
    double v1 = calc(m1);
    double v2 = calc(m2);
    if(v1 < v2) {
      right = m2;
    } else {
      left = m1;
    }
  }
  printf("%.10f\n", calc(left));
  
  return 0;
}