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 rep(i, n) for (int i = 0; i < (n); ++i)
#define mins(x, y) x = min(x, y)
#define maxs(x, y) x = max(x, y)

using namespace std;

const double inf=1e9;

struct D {
  vector<double> low, high;
  D(): low(3, inf), high(3, -inf) {}

  void add(double x, int v) {
    ++v;
    mins(low[v], x);
    maxs(high[v], x);
  }

  double length(double t) {
    double min_low = inf, max_high = -inf;
    rep(i, 3) {
      mins(min_low,  low[i] + i * t);
      maxs(max_high, high[i] + i * t);
    }
    return max_high - min_low;
  }

  vector<double> events() {
    vector<double> ts;
    rep(i, 3)rep(j,i) {
      {
        double t = (low[j] - low[i]) / (i - j);
        if (t > 0) ts.push_back(t);
      }

      {
        double t = (high[j] - high[i]) / (i - j);
        if (t > 0) ts.push_back(t);
      }
    }
    return ts;
  }
};

int main() {
  int n;
  cin >> n;
  D xd, yd;
  rep(i, n) {
    int x, y;
    char d;
    cin >> x >> y >> d;
    switch (d) {
      case 'L': xd.add(x, -1), yd.add(y, 0); break;
      case 'R': xd.add(x, 1), yd.add(y, 0); break;
      case 'U': xd.add(x, 0), yd.add(y, 1); break;
      case 'D': xd.add(x, 0), yd.add(y, -1); break;
    }
  }
  vector<double> ts;
  ts.push_back(0);
  ts.push_back(inf);
  {
    auto nts = xd.events();
    ts.insert(ts.end(), nts.begin(), nts.end());
  }
  {
    auto nts = yd.events();
    ts.insert(ts.end(), nts.begin(), nts.end());
  }

  sort(ts.begin(), ts.end());
  double ans = 1e17;
  for (double t: ts) {
    double now = xd.length(t) * yd.length(t);
    mins(ans, now);
  }
  printf("%.10f\n", ans);
  return 0;

}