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:
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() { 
  ios::sync_with_stdio(false); cin.tie(0);
  const string dir = "RLUD";
  const int dx[] = {1, -1, 0, 0};
  const int dy[] = {0, 0, 1, -1};
  int n;
  cin >> n;
  vector<vector<long long>> x(4), y(4);
  for (int i = 0; i < n; i++) {
    long long xin, yin;
    cin >> xin >> yin;
    char c;
    cin >> c;
    int d = dir.find(c);
    x[d].push_back(xin * 2);
    y[d].push_back(yin * 2);
  }
  vector<long long> xs, ys;
  for (int d = 0; d < 4; d++) {
    if (x[d].size()) {
      xs.push_back(*max_element(x[d].begin(), x[d].end()));
      xs.push_back(*min_element(x[d].begin(), x[d].end()));
    }
    if (y[d].size()) {
      ys.push_back(*max_element(y[d].begin(), y[d].end()));
      ys.push_back(*min_element(y[d].begin(), y[d].end()));
    }
  }
  vector<long long> time_candi;
  for (vector<long long> v : {xs, ys}) {
    for (long long i : v) {
      for (long long j : v) {
        long long dif = abs(i - j);
        time_candi.push_back(dif / 2);
        time_candi.push_back(dif);
      }
    }
  }
  const long long inf = 1 << 30;
  long long ans = 1LL << 60;
  for (long long t : time_candi) {
    long long x_min = inf, x_max = -inf;
    long long y_min = inf, y_max = -inf;
    for (int d = 0; d < 4; d++) {
      for (long long xx : x[d]) {
        x_min = min(x_min, xx + t * dx[d]);
        x_max = max(x_max, xx + t * dx[d]);
      }
      for (long long yy : y[d]) {
        y_min = min(y_min, yy + t * dy[d]);
        y_max = max(y_max, yy + t * dy[d]);
      }
    }
    ans = min(ans, (x_max - x_min) * (y_max - y_min));
  }
  cout << setprecision(20) << fixed << 1. * ans / 4 << endl;
  return 0;
}