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 <algorithm>
#include <bitset>
#include <cassert>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<double, ll> state;
typedef vector<vector<ll>> matrix;

int x[100005];
int y[100005];
string d[100005];
int n;

double getArea(double t) {
  double xMax = -1e9;
  double xMin = 1e9;
  double yMax = -1e9;
  double yMin = 1e9;
  for(int i = 0; i < n; i++) {
    double xx = x[i];
    double yy = y[i];
    if(d[i] == "R") xx += t;
    if(d[i] == "L") xx -= t;
    if(d[i] == "U") yy += t;
    if(d[i] == "D") yy -= t;
    xMax = max(xMax, xx);
    xMin = min(xMin, xx);
    yMax = max(yMax, yy);
    yMin = min(yMin, yy);
  }
  return (xMax-xMin) * (yMax-yMin);
}

void solve() {
  cin >> n;
  for(int i = 0; i < n; i++) cin >> x[i] >> y[i] >> d[i];
  double lhs = 0;
  double rhs = 2e8;
  double ret = getArea(0);
  for(int qq = 0; qq < 200; qq++) {
    double a = (2*lhs+rhs)/3;
    double b = (2*rhs+lhs)/3;
    ret = min(ret, getArea(a));
    ret = min(ret, getArea(b));
    if(getArea(a) <= getArea(b)) rhs = b;
    else lhs = a;
  }
  cout << fixed << setprecision(10) << ret << "\n";
}

void casesolve() {
  /*
  int t;
  cin >> t;
  for(int i = 1; i <= t; i++) {
    cout << "Case #" << i << ": ";
    solve();
  }
  */
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);
  solve();
  // casesolve();
}