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;

typedef long long ll;

const int N(2e5 + 9);
const double inf(1e18);

int T, n, m, ans;

struct A
{
    double x, y;
    char d;
}a[N];

double calc(double t)
{
    double minx = inf, maxx = -inf, miny = inf, maxy = -inf;
    for (int i = 0; i < n; i++)
    {
        double x = a[i].x;
        double y = a[i].y;
        if (a[i].d == 'L') x -= t;
        else if (a[i].d == 'R') x += t;
        else if (a[i].d == 'U') y += t;
        else if (a[i].d == 'D') y -= t;
        minx = min(minx, x);
        miny = min(miny, y);
        maxx = max(maxx, x);
        maxy = max(maxy, y);
       /* minx = min(minx, t * (a[i].d == 'L' ? -1 : a[i].d == 'R') + a[i].x);
        maxx = max(maxx, t * (a[i].d == 'L' ? -1 : a[i].d == 'R') + a[i].x);
        miny = min(miny, t * (a[i].d == 'D' ? -1 : a[i].d == 'U') + a[i].y);
        maxy = max(maxy, t * (a[i].d == 'D' ? -1 : a[i].d == 'U') + a[i].y);*/

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

int main()
{
    cin>>n;
    for (int i = 0; i < n; i++)
    {
        cin>>a[i].x>>a[i].y>>a[i].d;
    }
    double l = 0, r = 1e9, midl, midr;
    for (int i = 0; i< 150; i++)
    {
        midl = (2 * l + r) / 3;
        midr = (2 * r + l) / 3;
        if (calc(midl) < calc(midr))
        {
            r = midr;
        }
        else l = midl;
    }
     cout << fixed << setprecision(9) <<min(calc(0), calc(l)) << endl;
    return 0;
}