#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<int(n);++i)

typedef complex<double> Point;
typedef vector<Point> VP;
#define X real()
#define Y imag()

int main(void){
    int n;
    cin >> n;
    VP ps(n);
    vector<double> dx(n), dy(n);
    rep(i,n) {
        double x,y;
        char ch;
        cin >> x >> y >> ch;
        ps[i] = Point(x,y);
        if (ch == 'R') dx[i] = +1.0;
        if (ch == 'L') dx[i] = -1.0;
        if (ch == 'U') dy[i] = +1.0;
        if (ch == 'D') dy[i] = -1.0;
    }
    double l = 0, r = 1e9;
    auto area = [&](double t) {
        double minX = 1e9, minY = 1e9, maxX = -(1e9), maxY = -(1e9);
        rep(i,n){
            double x = ps[i].X + dx[i]*t;
            double y = ps[i].Y + dy[i]*t;
            minX = min(minX, x);
            minY = min(minY, y);
            maxX = max(maxX, x);
            maxY = max(maxY, y);
        }
        return (maxX - minX) * (maxY - minY);
    };
    rep(loop,500){
        double m1 = (l*2.0 + r)/3.0;
        double m2 = (l + r*2.0)/3.0;
        if (area(m1) <= area(m2)) {
            r = m2;
        } else {
            l = m1;
        }
    }
    printf("%.15f\n", area((l+r)/2.0));
    return 0;
}

### Prompt

Please create a solution in Cpp 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;
#define rep(i,n) for(int i=0;i<int(n);++i)

typedef complex<double> Point;
typedef vector<Point> VP;
#define X real()
#define Y imag()

int main(void){
    int n;
    cin >> n;
    VP ps(n);
    vector<double> dx(n), dy(n);
    rep(i,n) {
        double x,y;
        char ch;
        cin >> x >> y >> ch;
        ps[i] = Point(x,y);
        if (ch == 'R') dx[i] = +1.0;
        if (ch == 'L') dx[i] = -1.0;
        if (ch == 'U') dy[i] = +1.0;
        if (ch == 'D') dy[i] = -1.0;
    }
    double l = 0, r = 1e9;
    auto area = [&](double t) {
        double minX = 1e9, minY = 1e9, maxX = -(1e9), maxY = -(1e9);
        rep(i,n){
            double x = ps[i].X + dx[i]*t;
            double y = ps[i].Y + dy[i]*t;
            minX = min(minX, x);
            minY = min(minY, y);
            maxX = max(maxX, x);
            maxY = max(maxY, y);
        }
        return (maxX - minX) * (maxY - minY);
    };
    rep(loop,500){
        double m1 = (l*2.0 + r)/3.0;
        double m2 = (l + r*2.0)/3.0;
        if (area(m1) <= area(m2)) {
            r = m2;
        } else {
            l = m1;
        }
    }
    printf("%.15f\n", area((l+r)/2.0));
    return 0;
}

```