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 <string>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
#include <random>

using namespace std;


int main()
{
	int N;
	vector<long long> Xs;
	vector<long long> Ys;
	vector<int> DXs;
	vector<int> DYs;

	cin >> N;	

	long long ymax = -2e9;
	long long xmax = -2e9;
	long long ymin = 2e9;
	long long xmin = 2e9;

	long long x, y;
	char c;
	for (int ni = 0; ni < N; ni++){
		cin >> x >> y >> c;
		x *= 2;
		y *= 2;
		Xs.push_back(x);
		Ys.push_back(y);
		if (c == 'U'){
			DXs.push_back(0);
			DYs.push_back(1);
		}else if(c == 'D'){
			DXs.push_back(0);
			DYs.push_back(-1);
		}else if(c == 'R'){
			DXs.push_back(1);
			DYs.push_back(0);
		}else if(c == 'L'){
			DXs.push_back(-1);
			DYs.push_back(0);
		}
		ymax = max(ymax, y);
		xmax = max(xmax, x);
		ymin = min(ymin, y);
		xmin = min(xmin, x);
	}

	long long A = (ymax - ymin) * (xmax - xmin);
	long long minA = A;

	int cnt = 0;
	long long w,h;
	h = (ymax - ymin);
	w = (xmax - xmin);

	while(cnt < 1e3){
		ymax = -2e9;
		xmax = -2e9;
		ymin = 2e9;
		xmin = 2e9;
		for(int ni = 0; ni < N; ni++){
			Xs[ni] += DXs[ni];
			Ys[ni] += DYs[ni];
			ymax = max(ymax, Ys[ni]);
			xmax = max(xmax, Xs[ni]);
			ymin = min(ymin, Ys[ni]);
			xmin = min(xmin, Xs[ni]);
		}
		A = (ymax - ymin) * (xmax - xmin);
		if (A < minA) {
			minA = A;
		}

		if ( (ymax - ymin) >= h && (xmax - xmin) >= w){
			break;
		}

		h = (ymax - ymin);
		w = (xmax - xmin);

	}

	printf("%f",(double)(minA/4.0));

	return 0;
}