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 ll = long long;
using ii = pair<int, int>;
using ld = long double;

struct pt {
  ll x, y;
  int dx, dy;

  pt() {}
  pt(ll x, ll y) {}
  pt(ll x, ll y, char d): x(x), y(y) {
    if (d == 'R') dx = 1, dy = 0;
    else if (d == 'U') dx = 0, dy = 1;
    else if (d == 'L') dx = -1, dy = 0;
    else dx = 0, dy = -1;
  }
};

int n;
vector<pt> v;

ld f(ld t) {
  ld x[2] = {LLONG_MAX, LLONG_MIN}, y[2] = {LLONG_MAX, LLONG_MIN};
  for (int i = 0; i < n; i++) {
    x[0] = min(.0L + x[0], v[i].x + v[i].dx * t);
    x[1] = max(.0L + x[1], v[i].x + v[i].dx * t);

    y[0] = min(.0L + y[0], v[i].y + v[i].dy * t);
    y[1] = max(.0L + y[1], v[i].y + v[i].dy * t);
  }

  return (x[1] - x[0]) * (y[1] - y[0]);
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  cin >> n;

  v.reserve(n);

  int x, y;
  char d;
  for (int i = 0; i < n; i++) {
    cin >> x >> y >> d;

    v.push_back(pt(x, y, d));
  }

  ld lower = 0, upper = 1LL << 35;
  ld m1, m2;
  for (int i = 0; i < 500; i++) {
    m1 = lower + (upper - lower) / 3;
    m2 = upper - (upper - lower) / 3;

    if (f(m1) < f(m2)) upper = m2;
    else lower = m1;
  }

  cout << fixed << setprecision(12) << f(m1) << endl;
  
  return 0;
}