#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

ll N;
map<char, map<ll, set<double>>> Vs;
double R = 1e20;

void update(double t) {
  t = abs(t);
  double res = 1;
  for (char xy : {'x', 'y'}) {
    set<double> avs;
    for (const auto &d_vs : Vs[xy]) {
      ll d = d_vs.first;
      const auto &vs = d_vs.second;
      avs.insert(*vs.begin() + d * t);
      avs.insert(*vs.rbegin() + d * t);
    }
    res *= *avs.rbegin() - *avs.begin();
  }
  R = min(R, res);
}
int main() {
  cin >> N;
  for (ll i = 0; i < N; ++i) {
    double x, y;
    char d;
    cin >> x >> y >> d;
    Vs['x'][d == 'R' ? 1 : d == 'L' ? -1 : 0].insert(x);
    Vs['y'][d == 'U' ? 1 : d == 'D' ? -1 : 0].insert(y);
  }
  update(0);
  for (char xy : {'x', 'y'}) {
    for (ll d : {0, 1, -1}) {
      if (!Vs[xy].count(d)) continue;
      for (ll d2 : {0, 1, -1}) {
        if (d >= d2) continue;
        if (!Vs[xy].count(d2)) continue;
        const auto &vs = Vs[xy][d], &vs2 = Vs[xy][d2];
        update((*vs2.begin() - *vs.begin()) / (d - d2));
        update((*vs2.rbegin() - *vs.rbegin()) / (d - d2));
      }
    }
  }
  cout << setprecision(10) << R << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;

typedef long long ll;

ll N;
map<char, map<ll, set<double>>> Vs;
double R = 1e20;

void update(double t) {
  t = abs(t);
  double res = 1;
  for (char xy : {'x', 'y'}) {
    set<double> avs;
    for (const auto &d_vs : Vs[xy]) {
      ll d = d_vs.first;
      const auto &vs = d_vs.second;
      avs.insert(*vs.begin() + d * t);
      avs.insert(*vs.rbegin() + d * t);
    }
    res *= *avs.rbegin() - *avs.begin();
  }
  R = min(R, res);
}
int main() {
  cin >> N;
  for (ll i = 0; i < N; ++i) {
    double x, y;
    char d;
    cin >> x >> y >> d;
    Vs['x'][d == 'R' ? 1 : d == 'L' ? -1 : 0].insert(x);
    Vs['y'][d == 'U' ? 1 : d == 'D' ? -1 : 0].insert(y);
  }
  update(0);
  for (char xy : {'x', 'y'}) {
    for (ll d : {0, 1, -1}) {
      if (!Vs[xy].count(d)) continue;
      for (ll d2 : {0, 1, -1}) {
        if (d >= d2) continue;
        if (!Vs[xy].count(d2)) continue;
        const auto &vs = Vs[xy][d], &vs2 = Vs[xy][d2];
        update((*vs2.begin() - *vs.begin()) / (d - d2));
        update((*vs2.rbegin() - *vs.rbegin()) / (d - d2));
      }
    }
  }
  cout << setprecision(10) << R << endl;
  return 0;
}

```