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>

#define all(x) x.begin(), x.end()

using namespace std;

typedef long long ll;
typedef vector<ll> dlist;
typedef vector<dlist> matrix;

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cout << setprecision(10) << fixed;

    int n;
    cin >> n;

    double vx[2], vy[2];
    vector<char> d(n);
    vector<int> x(n), y(n);
    for(int i = 0; i < n; ++i)
        cin >> x[i] >> y[i] >> d[i];

    auto move = [&](int idx, double t){
        double a = 1.0 * x[idx] + (d[idx] == 'R') * t - (d[idx] == 'L') * t;
        double b = 1.0 * y[idx] + (d[idx] == 'U') * t - (d[idx] == 'D') * t;
        return pair<double, double>(a, b);
    };

    auto calc = [&](double t){
        auto v = move(0, t);
        vx[0] = vx[1] = v.first;
        vy[0] = vy[1] = v.second;
        for(int i= 0; i < n; ++i){
            v = move(i, t);
            vx[0] = min(vx[0], v.first);
            vx[1] = max(vx[1], v.first);
            vy[0] = min(vy[0], v.second);
            vy[1] = max(vy[1], v.second);
        }
        return (vx[1] - vx[0]) * (vy[1] - vy[0]);
    };

    double lo = 0, hi = 1e11, m1, m2, step, v1, v2;
    for(int i = 0; i <= 150; ++i){
        step = (hi - lo) / 3;
        m1 = lo + step;
        m2 = hi - step;

        v1 = calc(m1);
        v2 = calc(m2); 
        if(v1 < v2)
            hi = m2;
        else
            lo = m1;
    }
    cout << calc(lo) << "\n";

    return 0;
}//RUL0