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;

const int MAX = 2e5 + 5;
int n;
pair <int, int> a[MAX];
char d[MAX];

double f(double t) {
	double mnX = 1e10, mnY = 1e10;
	double mxX = -1e10, mxY = -1e10;

	for(int i = 1; i <= n; i++) {
		double X = a[i].first;
		double Y = a[i].second;

		if(d[i] == 'U')
			Y += t;

		else if(d[i] == 'R')
			X += t;

		else if(d[i] == 'D')
			Y -= t;

		else X -= t;

		mnX = min(mnX, X);
		mxX = max(mxX, X);

		mnY = min(mnY, Y);
		mxY = max(mxY, Y);
	}

	return (mxX - mnX) * (mxY - mnY);
}

int main() {
	scanf("%d", &n);

	for(int i = 1; i <= n; i++) {
		int x, y; char s[5];
		scanf("%d%d%s", &x, &y, s);
		a[i] = { x, y };
		d[i] = s[0];
	}

	double st = 0, nd = 6e8;

	for(int i = 1; i <= 200; i++) {
		double m1 = st + (nd - st) / 3.0;
		double m2 = nd - (nd - st) / 3.0;

		if(f(m1) < f(m2))
			nd = m2;

		else st = m1;
	}

	double opt = (st + nd) / 2.0;

	printf("%.10lf\n", f(opt));
	return 0;
}