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;
#define all(v) ((v).begin()),((v).end())
#define ll long long
#define fastIO cout << fixed << setprecision(12), ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
double const EPS = 1e-12, PI = acos(-1);
const int N = 2e5 + 9, M = 5e3 + 9, OO = 1e9 + 7, MOD = 1e9 + 7;
const ll inf = 1e18;

ll n, x[N], y[N];
char d[N];

long double f(long double val) {
  long double minX = inf, minY = inf, maxX = -inf, maxY = -inf;
  for (int i = 0; i < n; ++i) {
    ll dirX = 0, dirY = 0;
    if(d[i] == 'U')
      dirY = 1;
    else if(d[i] == 'D')
      dirY = -1;
    else if(d[i] == 'R')
      dirX = 1;
    else
      dirX = -1;
    minX = min(minX, 1.0 * x[i] + 1.0 * dirX * val);
    minY = min(minY, 1.0 * y[i] + 1.0 * dirY * val);
    maxX = max(maxX, 1.0 * x[i] + 1.0 * dirX * val);
    maxY = max(maxY, 1.0 * y[i] + 1.0 * dirY * val);
  }
  return (maxX - minX) * (maxY - minY);
}

int main() {
  fastIO;
#ifdef LOCAL
  freopen("input.in", "rt", stdin);
#endif
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i] >> d[i];
  }
  long double start = 0, end = 1e9 + 5;
  for (int i = 0; i < 750; ++i) {
    long double mid1 = start + (end - start) / 3;
    long double mid2 = end - (end - start) / 3;
    long double f1 = f(mid1);
    long double f2 = f(mid2);
    if(f1 < f2)
      end = mid2;
    else
      start = mid1;
  }
  cout << f(start);
  return 0;
}