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 <algorithm>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>

using namespace std;

typedef long long ll;

#define fi first
#define se second
#define mp make_pair
#define pb push_back

#define range(i, a, b, s) for (int i = (a), __##i = (b); i < __##i; i += (s))
#define down(i, a, b, s) for (int i = (a), __##i = (b); __##i < i; i += (s))

const int maxn = 1e5 + 2;

char c[maxn];
int n, x[maxn], y[maxn], dx[maxn], dy[maxn];

long double solve(long double t) {
  long double lx = +1e8, ly = +1e8;
  long double rx = -1e8, ry = -1e8;
  range(i, 0, n, 1) {
    long double xx = x[i] + dx[i] * t;
    long double yy = y[i] + dy[i] * t;
    lx = min(lx, xx);
    ly = min(ly, yy);
    rx = max(rx, xx);
    ry = max(ry, yy);
  }
  return (rx - lx) * (ry - ly);
}

int main(int argc, char const *argv[]) {
  // freopen("F.txt", "r", stdin);

  while (cin >> n) {
    range(i, 0, n, 1) {
      cin >> x[i] >> y[i] >> c[i];
      if (c[i] == 'R') dx[i] = +1, dy[i] = 0;
      if (c[i] == 'L') dx[i] = -1, dy[i] = 0;
      if (c[i] == 'U') dx[i] = 0, dy[i] = +1;
      if (c[i] == 'D') dx[i] = 0, dy[i] = -1;
    }
    long double l = 0, r = 1e8;
    range(_, 0, 300, 1) {
      if (solve((l + l + r) / 3) < solve((l + r + r) / 3))
        r = (l + r + r) / 3;
      else
        l = (l + l + r) / 3;
    }
    cout << fixed << setprecision(10) << solve(l) << endl;
  }
  return 0;
}