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 Int = long long;
constexpr double mv[2][4] = {
    { +1, -1, 0, 0 }, { 0, 0, +1, -1 }
};
struct Data { Int axis[2], dir; };
double range(const vector<Data>& D, double t, int axis) {
    double M = -1e9, m = 1e9;
    for (auto &d : D) {
        double x = d.axis[axis] + t * mv[axis][d.dir];
        M = max(M, x);
        m = min(m, x);
    }
    return M - m;
}
double slope(const vector<Data>& D, double t, int axis) {
    return (range(D, t+0.1, axis) - range(D, t, axis)) / 0.1;
}
double func(const vector<Data>& D, double t) {
    return range(D, t, 0) * range(D, t, 1);
}
int main()
{
    int N; cin >> N;
    vector<Data> D(N);
    for (int i = 0; i < N; i++) {
        cin >> D[i].axis[0] >> D[i].axis[1];
        char c; cin >> c;
        D[i].dir = (c == 'R' ? 0 : c == 'L' ? 1 : c == 'U' ? 2 : 3);
    }
    vector<double> T;
    T.push_back(0);
    for (int axis = 0; axis < 2; axis++) {
        double d = slope(D, 0.25, axis);
        Int curr = 0;
        while (curr < 2e8) {
            Int ok = 2e8, ng = curr;
            while (ok - ng > 1) {
                Int mid = (ok + ng) / 2;
                double t = mid / 2.0 + 0.25;
                (slope(D, t, axis) > d + 0.5 ? ok : ng) = mid;
            }
            T.push_back(ok / 2.0);
            curr = ok;
            d = slope(D, ok / 2.0 + 0.25, axis);
        }
    }
    double ans = 1e18;
    for (auto t : T) {
        ans = min(ans, func(D, t));
    }
    cout << fixed << setprecision(12) << ans << '\n';
    return 0;
}