#include <bits/stdc++.h>
using namespace std;
const double oo = 1e18;
 
int n, x[100100], y[100100];
char d[100100];
 
double calc(double t)
{
  double xMax = -oo, xMin = oo, yMax = -oo, yMin = oo;
  for (int i = 0; i < n; i++)
  {
    double curX = x[i], curY = y[i];
    if (d[i] == 'U') curY += t;
    else if (d[i] == 'D') curY -= t;
    else if (d[i] == 'R') curX += t;
    else curX -= t;
    xMax = max(xMax, curX);
    xMin = min(xMin, curX);
    yMax = max(yMax, curY);
    yMin = min(yMin, curY);
  }
  return (xMax - xMin) * (yMax - yMin);
}
 
int main()
{
  ios::sync_with_stdio(0);
  
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> x[i] >> y[i] >> d[i];
 
  double ans = calc(0), low = 0, high = 1e9;
  for (int turn = 0; turn < 150; turn++)
  {
    double mid1 = (low * 2 + high) / 3, v1 = calc(mid1);
    double mid2 = (low + high * 2) / 3, v2 = calc(mid2);
    if (v1 < v2) high = mid2;
    else low = mid1;
  }
 
  ans = min(ans, calc(low));
  cout << fixed << setprecision(9) << ans << endl;
}
### Prompt

Generate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double oo = 1e18;
 
int n, x[100100], y[100100];
char d[100100];
 
double calc(double t)
{
  double xMax = -oo, xMin = oo, yMax = -oo, yMin = oo;
  for (int i = 0; i < n; i++)
  {
    double curX = x[i], curY = y[i];
    if (d[i] == 'U') curY += t;
    else if (d[i] == 'D') curY -= t;
    else if (d[i] == 'R') curX += t;
    else curX -= t;
    xMax = max(xMax, curX);
    xMin = min(xMin, curX);
    yMax = max(yMax, curY);
    yMin = min(yMin, curY);
  }
  return (xMax - xMin) * (yMax - yMin);
}
 
int main()
{
  ios::sync_with_stdio(0);
  
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> x[i] >> y[i] >> d[i];
 
  double ans = calc(0), low = 0, high = 1e9;
  for (int turn = 0; turn < 150; turn++)
  {
    double mid1 = (low * 2 + high) / 3, v1 = calc(mid1);
    double mid2 = (low + high * 2) / 3, v2 = calc(mid2);
    if (v1 < v2) high = mid2;
    else low = mid1;
  }
 
  ans = min(ans, calc(low));
  cout << fixed << setprecision(9) << ans << endl;
}
```