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 <cstdio>
#include <cstdint>
#include <vector>
#include <algorithm>
#include <string>
#include <utility>
#include <numeric>
#include <queue>
#include <deque>
#include <tuple>
#include <map>
#include <set>

double inf = 1e12;

double f(double t, std::vector<std::tuple<intmax_t, intmax_t, char>> const& xyd) {
  double xmax = -inf;
  double xmin = inf;
  double ymax = -inf;
  double ymin = inf;

  for (auto const& p: xyd) {
    intmax_t xi, yi;
    char d;
    std::tie(xi, yi, d) = p;
    double x = xi;
    double y = yi;

    if (d == 'R') {
      x += t;
    } else if (d == 'L') {
      x -= t;
    } else if (d == 'U') {
      y += t;
    } else if (d == 'D') {
      y -= t;
    }

    xmax = std::max(xmax, x);
    xmin = std::min(xmin, x);
    ymax = std::max(ymax, y);
    ymin = std::min(ymin, y);
  }

  return (xmax-xmin) * (ymax-ymin);
}

int main() {
  size_t N;
  scanf("%zu", &N);

  std::vector<std::tuple<intmax_t, intmax_t, char>> xyd;
  xyd.reserve(N);
  for (size_t i = 0; i < N; ++i) {
    intmax_t x, y;
    char d;
    scanf("%jd %jd %c", &x, &y, &d);
    xyd.emplace_back(x, y, d);
  }

  double lb = 0;
  double ub = 1e9;
  for (int i = 0; i < 200; ++i) {
    double midl = (lb+lb+ub) / 3;
    double midu = (lb+ub+ub) / 3;
    if (f(midl, xyd) > f(midu, xyd)) {
      lb = midl;
    } else {
      ub = midu;
    }
  }

  printf("%.12f\n", std::min(f(lb, xyd), f(ub, xyd)));
}