#include <bits/stdc++.h>

using namespace std;

tuple<int,int,int> points[(int)1e5+10];
int movi[] = {0,0,-1,1};
int movj[] = {1,-1,0,0};
int n;

double solve(double t){
    double minx = 1e12;
    double maxx = -1e12;
    double miny = 1e12;
    double maxy = -1e12;

    for(int i=0;i<n;++i){
        double x = get<0>(points[i]);
        double y = get<1>(points[i]);
        int dir = get<2>(points[i]);
        x = x + t * movi[dir];
        y = y + t * movj[dir];
        minx = min(minx, x);
        maxx = max(maxx, x);
        miny = min(miny, y);
        maxy = max(maxy, y);
    }

    return (maxx-minx) * (maxy-miny);
}

int main(){

    cout << setprecision(9) << fixed;

    ios_base::sync_with_stdio(0);
    cin.tie(0);

    map<char,int> change;
    change['U'] = 0;
    change['D'] = 1;
    change['L'] = 2;
    change['R'] = 3;

    cin >> n;

    int x, y;
    char dir;
    for(int i=0;i<n;++i){
        cin >> x >> y >> dir;
        points[i] = make_tuple(x,y,change[dir]);
    }

    double left = 0, right = 1e12;
    for(int i=0;i<200;++i){
        double m1 = left+(right-left)/3;
        double m2 = left+2*(right-left)/3;
        if(solve(m1) < solve(m2)) right = m2;
        else left = m1;
    }

    cout << solve(left) << '\n';

    return 0;
}

### Prompt

Generate a cpp solution 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;

tuple<int,int,int> points[(int)1e5+10];
int movi[] = {0,0,-1,1};
int movj[] = {1,-1,0,0};
int n;

double solve(double t){
    double minx = 1e12;
    double maxx = -1e12;
    double miny = 1e12;
    double maxy = -1e12;

    for(int i=0;i<n;++i){
        double x = get<0>(points[i]);
        double y = get<1>(points[i]);
        int dir = get<2>(points[i]);
        x = x + t * movi[dir];
        y = y + t * movj[dir];
        minx = min(minx, x);
        maxx = max(maxx, x);
        miny = min(miny, y);
        maxy = max(maxy, y);
    }

    return (maxx-minx) * (maxy-miny);
}

int main(){

    cout << setprecision(9) << fixed;

    ios_base::sync_with_stdio(0);
    cin.tie(0);

    map<char,int> change;
    change['U'] = 0;
    change['D'] = 1;
    change['L'] = 2;
    change['R'] = 3;

    cin >> n;

    int x, y;
    char dir;
    for(int i=0;i<n;++i){
        cin >> x >> y >> dir;
        points[i] = make_tuple(x,y,change[dir]);
    }

    double left = 0, right = 1e12;
    for(int i=0;i<200;++i){
        double m1 = left+(right-left)/3;
        double m2 = left+2*(right-left)/3;
        if(solve(m1) < solve(m2)) right = m2;
        else left = m1;
    }

    cout << solve(left) << '\n';

    return 0;
}

```