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 <bits/stdc++.h>

using namespace std;

vector<pair<double, double>> v;
vector<int> mov_x;
vector<int> mov_y;

double Area(double t)
{
    double max_x = v[0].first + t*mov_x[0];
    double max_y = v[0].second + t*mov_y[0];
    double min_x = max_x;
    double min_y = max_y;

    for(int i = 1; i < v.size(); i++){
        double a = v[i].first + t*mov_x[i];
        if(a > max_x)
            max_x = a;
        else if(a < min_x)
            min_x = a;

        a = v[i].second + t*mov_y[i];

        if(a > max_y)
            max_y = a;
        else if(a < min_y)
            min_y = a;
    }

    return abs((max_x - min_x) * (max_y - min_y));
}

int main() {

    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;

    v = vector<pair<double, double>>(n);
    mov_x = vector<int>(n, 0);
    mov_y = vector<int>(n, 0);

    for (int i = 0; i < n; ++i) {
        double a;
        double b;
        char l;
        cin >> a >> b >> l;

        v[i] = pair<double, double>(a, b);

        if(l == 'R')
            mov_x[i] = 1;
        else if(l == 'L')
            mov_x[i] = -1;
        else if(l == 'U')
            mov_y[i] = 1;
        else
            mov_y[i] = -1;
    }

    double L = 0;
    double R = pow(10, 16);

    for(int i = 0; i < 400; i++) {
        double m1 = L + (R - L)/3;
        double m2 = R - (R - L)/3;

        double a = Area(m1);
        double b = Area(m2);

        if(a > b)
            L = m1;
        else if(a < b)
            R = m2;
        else
            L = m1, R = m2;
    }

    cout.precision(11);
    cout << Area(L);

    return 0;
}