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;

using ll = long long;
using ld = long double;

int main() {
    cin.tie(0); ios::sync_with_stdio(false);
    cout << setprecision(10) << setiosflags(ios::fixed);

    int n;
    cin >> n;

    vector<double> x(n), y(n), dx(n), dy(n);
    for (int i = 0; i < n; ++i) {
        char d;
        cin >> x[i] >> y[i] >> d;

        if (d == 'R') dx[i] = 1;
        else if (d == 'L') dx[i] = -1;
        else if (d == 'U') dy[i] = 1;
        else if (d == 'D') dy[i] = -1;
    }

    const double INF = numeric_limits<double>::max();
    const double NINF = numeric_limits<double>::lowest();

    auto f = [&](const ld t) {
        pair<double, double> seg_x(INF, NINF), seg_y(INF, NINF);

        for (int i = 0; i < n; ++i) {
            double vx = x[i] + t * dx[i], vy = y[i] + t * dy[i];

            seg_x.first = min(seg_x.first, vx);
            seg_x.second = max(seg_x.second, vx);
            seg_y.first = min(seg_y.first, vy);
            seg_y.second = max(seg_y.second, vy);
        }
        return (seg_x.second - seg_x.first) * (seg_y.second - seg_y.first);
    };

    double min_v = f(0), lb = 0, ub = 1e9;
    for (int i = 0; i < 200; ++i) {
        const double t1 = (2.0 * lb + ub) / 3.0;
        const double t2 = (lb + 2.0 * ub) / 3.0;

        min_v = min(min_v, f(t1));
        min_v = min(min_v, f(t2));

        if (f(t1) > f(t2)) lb = t1;
        else ub = t2;
    }

    cout << min_v << endl;

    return 0;
}