#define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <limits>
#include <climits>
#include <cfloat>
#include <functional>
#include <iterator>
#include <memory>
#include <regex>
using namespace std;

const long long INF = LLONG_MAX / 2;

int n;
vector<long long> x, y;
vector<int> dx, dy;

double solve(double t)
{
    double xmin = DBL_MAX;
    double ymin = DBL_MAX;
    double xmax = -DBL_MAX;
    double ymax = -DBL_MAX;
    for(int i=0; i<n; ++i){
        double x2 = x[i] + dx[i] * t;
        double y2 = y[i] + dy[i] * t;
        xmin = min(xmin, x2);
        xmax = max(xmax, x2);
        ymin = min(ymin, y2);
        ymax = max(ymax, y2);
    }
    return (xmax - xmin) * (ymax - ymin);
}

int main()
{
    cin >> n;
    x.assign(n, 0);
    y.assign(n, 0);
    dx.assign(n, 0);
    dy.assign(n, 0);
    for(int i=0; i<n; ++i){
        char d;
        cin >> x[i] >> y[i] >> d;
        if(d == 'R')
            ++ dx[i];
        else if(d == 'L')
            -- dx[i];
        else if(d == 'U')
            ++ dy[i];
        else
            -- dy[i];
    }

    double left = 0.0;
    double right = 1.0e+10;
    for(int i=0; i<200; ++i){
        double mid1 = (left * 2.0 + right) / 3.0;
        double mid2 = (left + right * 2.0) / 3.0;
        if(solve(mid1) < solve(mid2))
            right = mid2;
        else
            left = mid1;
    }
    printf("%.10f\n", solve(left));

    return 0;
}

### 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
#define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <limits>
#include <climits>
#include <cfloat>
#include <functional>
#include <iterator>
#include <memory>
#include <regex>
using namespace std;

const long long INF = LLONG_MAX / 2;

int n;
vector<long long> x, y;
vector<int> dx, dy;

double solve(double t)
{
    double xmin = DBL_MAX;
    double ymin = DBL_MAX;
    double xmax = -DBL_MAX;
    double ymax = -DBL_MAX;
    for(int i=0; i<n; ++i){
        double x2 = x[i] + dx[i] * t;
        double y2 = y[i] + dy[i] * t;
        xmin = min(xmin, x2);
        xmax = max(xmax, x2);
        ymin = min(ymin, y2);
        ymax = max(ymax, y2);
    }
    return (xmax - xmin) * (ymax - ymin);
}

int main()
{
    cin >> n;
    x.assign(n, 0);
    y.assign(n, 0);
    dx.assign(n, 0);
    dy.assign(n, 0);
    for(int i=0; i<n; ++i){
        char d;
        cin >> x[i] >> y[i] >> d;
        if(d == 'R')
            ++ dx[i];
        else if(d == 'L')
            -- dx[i];
        else if(d == 'U')
            ++ dy[i];
        else
            -- dy[i];
    }

    double left = 0.0;
    double right = 1.0e+10;
    for(int i=0; i<200; ++i){
        double mid1 = (left * 2.0 + right) / 3.0;
        double mid2 = (left + right * 2.0) / 3.0;
        if(solve(mid1) < solve(mid2))
            right = mid2;
        else
            left = mid1;
    }
    printf("%.10f\n", solve(left));

    return 0;
}

```