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;

#define endl '\n'

typedef long long ll;
typedef pair<int, int> pii;

int len;
vector<char> D;
vector<double> X;
vector<double> Y;

vector<double> TX;
vector<double> TY;

void Time(double t){
	for (int i = 0; i < len; ++i) {
		if (D[i] == 'R') TX[i] = X[i] + t;
		else if (D[i] == 'L') TX[i] = X[i] - t;
		else{
			if (D[i] == 'U') TY[i] = Y[i] + t;
			if (D[i] == 'D') TY[i] = Y[i] - t;
			TX[i] = X[i];
			continue;
		}
		TY[i] = Y[i];
	}
}

double Ma(vector<double> vector){
	double maxi = vector[0];
	for (int i = 1; i < len; ++i) {
		maxi = max(vector[i],maxi);
	}
	return maxi;
}

double Mi(vector<double> vector){
	double maxi = vector[0];
	for (int i = 1; i < len; ++i) {
		maxi = min(vector[i],maxi);
	}
	return maxi;
}

double Ternary(){
	double L = 0;
	double R = pow(10,16);
	for (int i = 0; i < 400; ++i) {
		double m1 = L + (R - L)/3;
		double m2 = R - (R - L)/3;
		Time(m1);
		double d1 = (Ma(TX) - Mi(TX)) * (Ma(TY) - Mi(TY));
		Time(m2);
		double d2 = (Ma(TX) - Mi(TX)) * (Ma(TY) - Mi(TY));
		if (d1 > d2) L = m1;
		else if (d2 > d1) R = m2;
		else{
			L = m1;
			R = m2;
		}
	}
	Time(L);
//	cout << L << endl;
//	for (int j = 0; j < len; ++j) {
//		cout << TX[j] << " " << TY[j] << endl;
//	}
//	cout << endl;
//    cout << Ma(TX) << " " << Mi(TX) << " " << Ma(TY) << " " << Mi(TY) << endl;
	return (Ma(TX) - Mi(TX)) * (Ma(TY) - Mi(TY));
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	cin >> len;
	D = vector<char>(len), X = vector<double>(len),Y = vector<double>(len),TX = vector<double>(len),TY = vector<double>(len);
	
	for (int i = 0; i < len; ++i) {
		cin >> X[i] >> Y[i] >> D[i];
	}
	
	cout << fixed << setprecision(12) << Ternary();
	
	return 0;
}