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;
#define rep(i,n) for (int i=0;i<(n);i++)
#define chmin(a, b) a = min(a, b)
#define chmax(a, b) a = max(a, b)

const double inf = 1e9;

struct Data {
  vector<double> l, r;
  Data() : l(3, inf), r(3, -inf) {}
  void add(double x, int v) {
    v++;
    chmin(l[v], x);
    chmax(r[v], x);
  }
  vector<double> events() {
    vector<double> ts;
    rep(i, 3) rep(j, i) {
      {
        double t = (l[j] - l[i]) / (i - j);
        if (t > 0) ts.push_back(t);
      }
      {
        double t = (r[j] - r[i]) / (i - j);
        if (t > 0) ts.push_back(t);
      }
    }
    return ts;
  }
  double len(double t) {
    double nl = inf, nr = -inf;
    rep(v, 3) {
      chmin(nl, l[v]+v*t);
      chmax(nr, r[v]+v*t);
    }
    return nr - nl;
  }

  void show() {
    rep(i, 3) {
      cerr << "l = " << l[i] << ", r = " << r[i] << endl;
    }
  }
};

int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);

  int n;
  cin >> n;
  Data xd, yd;
  rep(i, n) {
    int x, y;
    char d;
    cin >> x >> y >> d;
    if (d == 'R') xd.add(x, +1), yd.add(y,  0);
    if (d == 'L') xd.add(x, -1), yd.add(y,  0);
    if (d == 'U') xd.add(x,  0), yd.add(y, +1);
    if (d == 'D') xd.add(x,  0), yd.add(y, -1);
  }

  vector<double> ts;
  ts.push_back(0);
  ts.push_back(inf);
  {
    auto xts = xd.events();
    ts.insert(ts.end(),xts.begin(), xts.end());
  }
  {
    auto yts = yd.events();
    ts.insert(ts.end(), yts.begin(), yts.end());
  }

  double ans = inf * inf;
  for (auto t : ts) {
    double area = xd.len(t) * yd.len(t);
    chmin(ans, area);
  }

  cout << fixed << setprecision(10) << ans << endl;
}