#include <bits/stdc++.h>
using namespace std;
typedef long double LD;
char s[2];
int x[100005], y[100005], d[100005];
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++){
        scanf("%d%d%s", &x[i], &y[i], s);
        if(s[0] == 'R') d[i] = 0;
        else if(s[0] == 'L') d[i] = 1;
        else if(s[0] == 'U') d[i] = 2;
        else d[i] = 3;
    }
    LD ans = 1e18;
    LD lo = 0, hi = 1e9;
    for(int i=0;i<200;i++){

        LD mid1 = lo+(hi-lo)/3;
        LD mx_x = -1e9, mn_x = 1e9, mx_y = -1e9, mn_y = 1e9;
        for(int i=1;i<=n;i++){
            mx_x = max(mx_x, (LD)x[i]+mid1*dir[d[i]][0]);
            mn_x = min(mn_x, (LD)x[i]+mid1*dir[d[i]][0]);

            mx_y = max(mx_y, (LD)y[i]+mid1*dir[d[i]][1]);
            mn_y = min(mn_y, (LD)y[i]+mid1*dir[d[i]][1]);
        }
        LD res1 = abs((mx_x-mn_x)*(mx_y-mn_y));

        LD mid2 = lo+2*(hi-lo)/3;
        mx_x = -1e9, mn_x = 1e9, mx_y = -1e9, mn_y = 1e9;
        for(int i=1;i<=n;i++){
            mx_x = max(mx_x, (LD)x[i]+mid2*dir[d[i]][0]);
            mn_x = min(mn_x, (LD)x[i]+mid2*dir[d[i]][0]);

            mx_y = max(mx_y, (LD)y[i]+mid2*dir[d[i]][1]);
            mn_y = min(mn_y, (LD)y[i]+mid2*dir[d[i]][1]);
        }
        LD res2 = abs((mx_x-mn_x)*(mx_y-mn_y));

        if(res1 < res2){
            ans = min(ans, res1);
            hi = mid2;
        }
        else{
            ans = min(ans, res2);
            lo = mid1;
        }
    }
    printf("%.20Lf\n", ans);
}
### 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;
typedef long double LD;
char s[2];
int x[100005], y[100005], d[100005];
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++){
        scanf("%d%d%s", &x[i], &y[i], s);
        if(s[0] == 'R') d[i] = 0;
        else if(s[0] == 'L') d[i] = 1;
        else if(s[0] == 'U') d[i] = 2;
        else d[i] = 3;
    }
    LD ans = 1e18;
    LD lo = 0, hi = 1e9;
    for(int i=0;i<200;i++){

        LD mid1 = lo+(hi-lo)/3;
        LD mx_x = -1e9, mn_x = 1e9, mx_y = -1e9, mn_y = 1e9;
        for(int i=1;i<=n;i++){
            mx_x = max(mx_x, (LD)x[i]+mid1*dir[d[i]][0]);
            mn_x = min(mn_x, (LD)x[i]+mid1*dir[d[i]][0]);

            mx_y = max(mx_y, (LD)y[i]+mid1*dir[d[i]][1]);
            mn_y = min(mn_y, (LD)y[i]+mid1*dir[d[i]][1]);
        }
        LD res1 = abs((mx_x-mn_x)*(mx_y-mn_y));

        LD mid2 = lo+2*(hi-lo)/3;
        mx_x = -1e9, mn_x = 1e9, mx_y = -1e9, mn_y = 1e9;
        for(int i=1;i<=n;i++){
            mx_x = max(mx_x, (LD)x[i]+mid2*dir[d[i]][0]);
            mn_x = min(mn_x, (LD)x[i]+mid2*dir[d[i]][0]);

            mx_y = max(mx_y, (LD)y[i]+mid2*dir[d[i]][1]);
            mn_y = min(mn_y, (LD)y[i]+mid2*dir[d[i]][1]);
        }
        LD res2 = abs((mx_x-mn_x)*(mx_y-mn_y));

        if(res1 < res2){
            ans = min(ans, res1);
            hi = mid2;
        }
        else{
            ans = min(ans, res2);
            lo = mid1;
        }
    }
    printf("%.20Lf\n", ans);
}
```