#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
#define rep(i,s,e) for(i64 (i) = (s);(i) < (e);(i)++)
#define rev(i,s,e) for(i64 (i) = (s);(i) --> (e);)
#define all(x) x.begin(),x.end()

int main() {
  i64 N;
  cin >> N;
  vector<double> X(N), Y(N);
  vector<char> D(N);
  map<char, pair<i64, i64>> mp;
  mp['R'] = {1, 0};
  mp['L'] = {-1, 0};
  mp['U'] = {0, 1};
  mp['D'] = {0, -1};
  rep(i,0,N) cin >> X[i] >> Y[i] >> D[i];

  auto f = [&](double t) {
    double xM = -1e18;
    double xm = 1e18;
    double yM = -1e18;
    double ym = 1e18;
    rep(i,0,N) {
      auto p = mp[D[i]];
      xM = max(xM, X[i] + p.first * t);
      xm = min(xm, X[i] + p.first * t);
      yM = max(yM, Y[i] + p.second * t);
      ym = min(ym, Y[i] + p.second * t);
    }
    return (xM - xm) * (yM - ym);
  };
  double b = 1e8;
  double a = 0;
  rep(i,0,200) {
    double c = (a * 2 + b) / 3;
    double d = (a + b * 2) / 3;
    if(f(c) < f(d)) {
      b = d;
    }
    else {
      a = c;
    }
  }
  cout << fixed << setprecision(10);
  cout << f(a) << endl;
}

### Prompt

Your task is to create 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;
using i64 = long long;
#define rep(i,s,e) for(i64 (i) = (s);(i) < (e);(i)++)
#define rev(i,s,e) for(i64 (i) = (s);(i) --> (e);)
#define all(x) x.begin(),x.end()

int main() {
  i64 N;
  cin >> N;
  vector<double> X(N), Y(N);
  vector<char> D(N);
  map<char, pair<i64, i64>> mp;
  mp['R'] = {1, 0};
  mp['L'] = {-1, 0};
  mp['U'] = {0, 1};
  mp['D'] = {0, -1};
  rep(i,0,N) cin >> X[i] >> Y[i] >> D[i];

  auto f = [&](double t) {
    double xM = -1e18;
    double xm = 1e18;
    double yM = -1e18;
    double ym = 1e18;
    rep(i,0,N) {
      auto p = mp[D[i]];
      xM = max(xM, X[i] + p.first * t);
      xm = min(xm, X[i] + p.first * t);
      yM = max(yM, Y[i] + p.second * t);
      ym = min(ym, Y[i] + p.second * t);
    }
    return (xM - xm) * (yM - ym);
  };
  double b = 1e8;
  double a = 0;
  rep(i,0,200) {
    double c = (a * 2 + b) / 3;
    double d = (a + b * 2) / 3;
    if(f(c) < f(d)) {
      b = d;
    }
    else {
      a = c;
    }
  }
  cout << fixed << setprecision(10);
  cout << f(a) << endl;
}

```