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 F first
#define S second
using namespace std;
typedef long long ll;
typedef long double ld;

const int mod = 1000000007;
const int max_n = 100011;

ld check(int n, int x[], int y[], char c[], ld t) {
  ld max1 = -mod, max2 = -mod, min1 = mod, min2 = mod;
  for (int i = 0; i < n; ++i) {
    if (c[i] == 'R') {
      max1 = max(max1, x[i] + t);
      min1 = min(min1, x[i] + t);
    }
    if (c[i] == 'L') {
      max1 = max(max1, x[i] - t);
      min1 = min(min1, x[i] - t);
    }
    if (c[i] == 'R' || c[i] == 'L') {
      max2 = max(max2, (ld)y[i]);
      min2 = min(min2, (ld)y[i]);
    }
    if (c[i] == 'U' || c[i] == 'D') {
      max1 = max(max1, (ld)x[i]);
      min1 = min(min1, (ld)x[i]);
    }
    if (c[i] == 'U') {
      max2 = max(max2, y[i] + t);
      min2 = min(min2, y[i] + t);
    }
    if (c[i] == 'D') {
      max2 = max(max2, y[i] - t);
      min2 = min(min2, y[i] - t);
    }
  }
  //cout << "check " << t << " = " << (max1 - min1) * (max2 - min2) << "   " << min1 << " " << max1 << " " << min2 << " " << max2 << endl;
  return (max1 - min1) * (max2 - min2);
}

int main() {
  int n, x[max_n], y[max_n];
  char c[max_n];
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d%d%*c%c", &x[i], &y[i], &c[i]);
    //cout << x[i] << " " << y [i] << " " << c[i] << endl;
  }
  ld l = 0, r = 1000000000;
  for (int i = 0; i < 400; ++i) {
    //cout << l << " " << r << endl;
    ld mid1 = (2 * l + r) / 3;
    ld mid2 = (l + 2 * r) / 3;
    if (check(n, x, y, c, mid1) < check(n, x, y, c, mid2)) {
      r = mid2;
    } else {
      l = mid1;
    }
  }
  printf("%.14f", double(check(n, x, y, c, l)));
}