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<iomanip>
#include<vector>
#include<string>
#define rep(i, start, end) for (int i = (int)start; i < (int)end; ++i)
#define rrep(i, start, end) for (int i = (int)start - 1; i >= end; --i)
#define iter(i, c) for (auto i = (c).begin(); i != (c).end(); ++i)
#define riter(i, c) for (auto i = (c).rbegin(); i != (c).rend(); ++i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
using namespace std;
using ll = long long;

const ll INF = 1LL << 50;

double f(const vector<double>& x, const vector<double>& y, const vector<char>& d, double t) {
    double max_x = -INF;
    double min_x = INF;
    double max_y = -INF;
    double min_y = INF;
    rep(i, 0, x.size()) {
        double now_x = x[i];
        double now_y = y[i];
        if (d[i] == 'U') now_y += t;
        else if (d[i] == 'D') now_y -= t;
        else if (d[i] == 'L') now_x -= t;
        else now_x += t;
        max_x = max(max_x, now_x);
        min_x = min(min_x, now_x);
        max_y = max(max_y, now_y);
        min_y = min(min_y, now_y);
    }
    return (max_x - min_x) * (max_y - min_y);
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    vector<double> X(N), Y(N);
    vector<char> D(N);
    rep(i, 0, N) cin >> X[i] >> Y[i] >> D[i];
    double l = 0.0, r = 1e10;
    rep(i, 0, 500) {
        double t1 = (2 * l + r) / 3;
        double t2 = (l + 2 * r) / 3;
        if (f(X, Y, D, t1) < f(X, Y, D, t2)) {
            r = t2;
        }
        else l = t1;
    }
    cout << fixed << setprecision(10) << f(X, Y, D, l) << endl;
    return 0;
}