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:
#pragma GCC optimize("O3", "unroll-loops")
#include <bits/stdc++.h>

using namespace std;
const int maxn = 1e5 + 10;
pair<double, double> a[maxn];
double dx[maxn], dy[maxn], ans, lb = 0, lbb, rbb, rb = 2e8;

int n;

char c;

double calc(double t) {
	double u, d, l, r;
	l = r = a[0].first + t * dx[0], u = d = a[0].second + t * dy[0];
	for (int i = 1; i < n; i++) {
		l = min(l, a[i].first + t * dx[i]);
		r = max(r, a[i].first + t * dx[i]);
		u = max(u, a[i].second + t * dy[i]);
		d = min(d, a[i].second + t * dy[i]);
	}
	return (u - d) * (r - l);
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i].first >> a[i].second >> c;
		if (c == 'U')
			dx[i] = 0, dy[i] = 1;
		else if (c == 'D')
			dx[i] = 0, dy[i] = -1;
		else if (c == 'R')
			dx[i] = 1, dy[i] = 0;
		else
			dx[i] = -1, dy[i] = 0;
	}
	ans = calc(0);
	for (int i = 0; i < 1000; i++) {
		lbb = (2 * lb + rb) / 3;
		rbb = (lb + 2 * rb) / 3;
		if (calc(lbb) < calc(rbb))
			rb = rbb;
		else
			lb = lbb;
	}
	cout << setprecision(10) << fixed << calc(lb) << '\n';
	return 0;
}