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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

int n;
long double x_p[100001], y_p[100001], x_d[100001], y_d[100001];

long double get_bound(long double mult){

    long double minix =  1e18;
    long double maxix = -1e18;

    long double miniy =  1e18;
    long double maxiy = -1e18;

    for(int i = 0; i < n; ++i){
        long double x = x_p[i] + x_d[i] * mult;
        long double y = y_p[i] + y_d[i] * mult;
        minix = min(minix, x);
        maxix = max(maxix, x);
        miniy = min(miniy, y);
        maxiy = max(maxiy, y);
    }

    return (maxix - minix) * (maxiy - miniy);

}

int main(){

    cin.sync_with_stdio(0); cin.tie(0); cout.tie(0);

    cin >> n;

    for(int i = 0; i < n; ++i){
        long double x, y;
        char d;
        cin >> x_p[i] >> y_p[i] >> d;
        if(d == 'U'){
            y_d[i] = 1;
        } else if(d == 'D'){
            y_d[i] = -1;
        } else if(d == 'R'){
            x_d[i] = 1;
        } else {
            x_d[i] = -1;
        }
    }

    long double left = 0, right = 1000000000;

    long double best = -1;

    for(int it = 0; it < 200; ++it){
        long double middle1 = left + (right - left) / 3;
        long double middle2 = right - (right - left) / 3;
        long double bound1 = get_bound(middle1);
        long double bound2 = get_bound(middle2);
        if(bound1 < bound2){
            right = middle2;
            if(best == -1)
                best = bound1;
            best = min(best, bound1);
        } else {
            left = middle1;
            if(best == -1)
                best = bound2;
            best = min(best, bound2);
        }
    }

    cout << fixed << setprecision(11) << best;

    return 0;
}