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;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n; cin >> n;
  V<long double> x(n), y(n);
  V<char> d(n);
  for (int i = 0; i < n; ++i) cin >> x[i] >> y[i] >> d[i];
  auto fn = [&](long double t) -> long double {
    long double x_min = 1e12, x_max = -1e12;
    long double y_min = 1e12, y_max = -1e12;
    for (int i = 0; i < n; ++i) {
      if (d[i] == 'R') {
        x_min = min(x_min, x[i] + t);
        x_max = max(x_max, x[i] + t);
      } else if (d[i] == 'L') {
        x_min = min(x_min, x[i] - t);
        x_max = max(x_max, x[i] - t);
      } else {
        x_min = min(x_min, x[i]);
        x_max = max(x_max, x[i]);
      }
      if (d[i] == 'U') {
        y_min = min(y_min, y[i] + t);
        y_max = max(y_max, y[i] + t);
      } else if (d[i] == 'D') {
        y_min = min(y_min, y[i] - t);
        y_max = max(y_max, y[i] - t);
      } else {
        y_min = min(y_min, y[i]);
        y_max = max(y_max, y[i]);
      }
    }
    return (x_max - x_min) * (y_max - y_min);
  };
  long double l = 0, r = 1e10;
  for (int _ = 0; _ < 200; ++_) {
    long double ml = (2 * l + r) / 3, mr = (l + 2 * r) / 3;
    if (fn(ml) < fn(mr)) r = mr;
    else l = ml;
  }
  cout << fixed << setprecision(15) << fn((l + r) / 2) << '\n';
}