#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < n; i++)
int n, a[2], x[2][3][2], s;
double ans;
char d;
inline double calc(double time){
    double ret = 1;
    rep(i,2) ret *= min(min((double)x[i][0][0], x[i][1][0] + time), x[i][2][0] - time) - max(max((double)x[i][0][1], x[i][1][1] + time), x[i][2][1] - time);
    return ret;
}
int main(){
    scanf("%d", &n);
    rep(i,2) rep(j,3) x[i][j][1] = -(x[i][j][0] = 1e8);
    rep(i,n){
        scanf("%d%d %c", a, a+1, &d);
        int t = d == 'U' || d == 'D';
        int k = d == 'L' || d == 'D';
        x[t^1][0][0] = min(x[t^1][0][0], a[t^1]);
        x[t^1][0][1] = max(x[t^1][0][1], a[t^1]);
        x[t][1+k][0] = min(x[t][1+k][0], a[t]);
        x[t][1+k][1] = max(x[t][1+k][1], a[t]);
    }
    ans = calc(0);
    rep(i,2) rep(j,2){
        double cand[3] = {(double)x[i][0][j] - x[i][1][j], (double)x[i][2][j] - x[i][0][j], ((double)x[i][2][j] - x[i][1][j])/2};
        rep(k,3) if(cand[k] > 0) ans = min(calc(cand[k]), ans);
    }
    printf("%0.10f\n", ans);
}
### Prompt

Your task is to create 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;
#define rep(i,n) for(int i = 0; i < n; i++)
int n, a[2], x[2][3][2], s;
double ans;
char d;
inline double calc(double time){
    double ret = 1;
    rep(i,2) ret *= min(min((double)x[i][0][0], x[i][1][0] + time), x[i][2][0] - time) - max(max((double)x[i][0][1], x[i][1][1] + time), x[i][2][1] - time);
    return ret;
}
int main(){
    scanf("%d", &n);
    rep(i,2) rep(j,3) x[i][j][1] = -(x[i][j][0] = 1e8);
    rep(i,n){
        scanf("%d%d %c", a, a+1, &d);
        int t = d == 'U' || d == 'D';
        int k = d == 'L' || d == 'D';
        x[t^1][0][0] = min(x[t^1][0][0], a[t^1]);
        x[t^1][0][1] = max(x[t^1][0][1], a[t^1]);
        x[t][1+k][0] = min(x[t][1+k][0], a[t]);
        x[t][1+k][1] = max(x[t][1+k][1], a[t]);
    }
    ans = calc(0);
    rep(i,2) rep(j,2){
        double cand[3] = {(double)x[i][0][j] - x[i][1][j], (double)x[i][2][j] - x[i][0][j], ((double)x[i][2][j] - x[i][1][j])/2};
        rep(k,3) if(cand[k] > 0) ans = min(calc(cand[k]), ans);
    }
    printf("%0.10f\n", ans);
}
```