#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

int n;
double x[200000],y[200000];
char d[200000][2];
int dx[]{1,-1,0,0},dy[]{0,0,1,-1};
map<char,int>dir;

double f(double t,int mode){
  double Min=-1,Max=-1;
  for(int i = 0; i < n; i++) {
    double X;
    if(mode == 0) X = x[i] + dx[dir[d[i][0]]] * t;
    else X = y[i] + dy[dir[d[i][0]]] * t;
    if(i == 0) Min = Max = X;
    else Min = min(Min, X), Max = max(Max, X);
  } return Max - Min;
}

int main() {
  dir['R'] = 0; dir['L'] = 1;
  dir['U'] = 2; dir['D'] = 3;
  cin >> n;
  rep(i, n) scanf("%lf%lf%s",&x[i],&y[i],d[i]);
  double l = 0, r = 1e15;
  for(int i = 0; i < 200; i++) {
    double l1 = (l + l + r) / 3;
    double l2 = (l + r + r) / 3;
    if(f(l1, 0) * f(l1, 1) < f(l2, 0) * f(l2, 1))
      r = l2; else l = l1;
  } printf("%.12lf\n", f(l, 0) * f(l, 1));
}

### Prompt

Create a solution in CPP for 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>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

int n;
double x[200000],y[200000];
char d[200000][2];
int dx[]{1,-1,0,0},dy[]{0,0,1,-1};
map<char,int>dir;

double f(double t,int mode){
  double Min=-1,Max=-1;
  for(int i = 0; i < n; i++) {
    double X;
    if(mode == 0) X = x[i] + dx[dir[d[i][0]]] * t;
    else X = y[i] + dy[dir[d[i][0]]] * t;
    if(i == 0) Min = Max = X;
    else Min = min(Min, X), Max = max(Max, X);
  } return Max - Min;
}

int main() {
  dir['R'] = 0; dir['L'] = 1;
  dir['U'] = 2; dir['D'] = 3;
  cin >> n;
  rep(i, n) scanf("%lf%lf%s",&x[i],&y[i],d[i]);
  double l = 0, r = 1e15;
  for(int i = 0; i < 200; i++) {
    double l1 = (l + l + r) / 3;
    double l2 = (l + r + r) / 3;
    if(f(l1, 0) * f(l1, 1) < f(l2, 0) * f(l2, 1))
      r = l2; else l = l1;
  } printf("%.12lf\n", f(l, 0) * f(l, 1));
}

```