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 <vector>
#include <algorithm>
#include <functional>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<long long> x(N), y(N);
    vector<char> d(N);
    for (int i=0; i<N; i++)
        cin>>x[i]>>y[i]>>d[i];

    for (int i=0; i<N; i++)
    {
        x[i] *= 2;
        y[i] *= 2;
    }

    long long oo = 0x10000000LL;

    auto fx = [&](long long t) -> long long
    {
        long long mn = oo;
        long long mx = -oo;
        for (int i=0; i<N; i++)
        {
            long long a = x[i] + (d[i]=='R' ? t : d[i]=='L' ? -t : 0);
            mn = min(mn, a);
            mx = max(mx, a);
        }
        return mx-mn;
    };
    auto fy = [&](long long t) -> long long
    {
        long long mn = oo;
        long long mx = -oo;
        for (int i=0; i<N; i++)
        {
            long long a = y[i] + (d[i]=='U' ? t : d[i]=='D' ? -t : 0);
            mn = min(mn, a);
            mx = max(mx, a);
        }
        return mx-mn;
    };

    auto flex = [&](function<long long(long long)> f) -> vector<long long> 
    {
        long long s[] = {-2, -1, 0, 1};
        long long p = 0;
        vector<long long> r;
        for (int i=0; i<4; i++)
        {
            if (f(p+1)-f(p)==s[i])
            {
                for (long long b=oo; b>0; b/=2)
                    if (f(p+b+1)-f(p+b)==s[i])
                        p += b;
                p++;
            }
            r.push_back(p);
        }
        return r;
    };

    long long ans = fx(0)*fy(0);
    for (long long t: flex(fx))
        ans = min(ans, fx(t)*fy(t));
    for (long long t: flex(fy))
        ans = min(ans, fx(t)*fy(t));

    printf("%.2f\n", ans*.25);
}