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;

template <typename T=double>
pair<T,T> ternary(T l, T r, function<T(T)> f, bool V=true) {
    int round = 400;
    while (round--) {
        T m0 = l + (r-l)/3;
        T m1 = r - (r-l)/3;
        T f0 = f(m0);
        T f1 = f(m1);

        if ((f0 < f1) ^ V) {
            l = m0;
        } else {
            r = m1;
        }
    }
    return {l, f(l)};
}

void solve() {
    int n; cin >> n;
    vector<double> x(n), y(n);
    vector<int> sx(n, 0), sy(n, 0);
    for (int i = 0; i < n; i++) {
        char d;
        cin >> x[i] >> y[i] >> d;
        if (d == 'L') sx[i] = -1;
        if (d == 'R') sx[i] = +1;
        if (d == 'D') sy[i] = -1;
        if (d == 'U') sy[i] = +1;
    }

    double _, res;
    tie(_, res) = ternary<double>(0.0, 1e9, [&](double t){
            double INF = 1e18;
            double mx = INF, my = INF, Mx = -INF, My = -INF;
            for (int i = 0; i < n; i++) {
                double nx = x[i] + t * sx[i];
                double ny = y[i] + t * sy[i];
                mx = min(mx, nx);
                Mx = max(Mx, nx);
                my = min(my, ny);
                My = max(My, ny);
            }
            return (Mx-mx)*(My-my);
            });
    cout << fixed << setprecision(10) << res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}