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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <cmath>

#define int long long
//#define double long double

using namespace std;

long long n, m;
double ans = 1e18;
long long x[500000], y[500000];
char c[500000];

double get(double t)
{
    double minx, maxx, miny, maxy;
    for (int i = 0; i < n; i++)
    {
        double a = x[i], b = y[i];
        if (c[i] == 'D')
            b -= t;
        if (c[i] == 'U')
            b += t;
        if (c[i] == 'R')
            a += t;
        if (c[i] == 'L')
            a -= t;
        if (i == 0)
        {
            minx = a, miny = b;
            maxx = a, maxy = b;
        }
        else
        {
            minx = min(minx, a);
            miny = min(miny, b);
            maxy = max(maxy, b);
            maxx = max(maxx, a);
        }
    }
    return (maxx - minx) * (maxy - miny);
}

signed main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
        cin >> x[i] >> y[i] >> c[i];
    double l = 0, r = 6e8;
    for (int i = 0; i < 1000; i++)
    {
        //printf("%.9f %.9f\n", l, r);
        double m1 = l + (r - l) * 1 / 3;
        double m2 = l + (r - l) * 2 / 3;
        double s1 = get(m1);
        double s2 = get(m2);
        if (s1 > s2)
            l = m1;
        else
            r = m2;
    }
    l = round(2 * l) / 2;
    //printf("%.9f\n", l);
    printf("%.9f", get(l));
}