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;
using Int = int_fast64_t;

Int n;
typedef pair<Int, Int> P;
vector<P> x, y;
constexpr double EPS = 1e-14;

double func(double t){
	double xmi = 1e10, ymi = 1e10, xma = -1e10, yma = -1e10;
	for(Int i=0; i<n; ++i){
		xmi = min(xmi, x[i].first+t*x[i].second);
		xma = max(xma, x[i].first+t*x[i].second);
		ymi = min(ymi, y[i].first+t*y[i].second);
		yma = max(yma, y[i].first+t*y[i].second);
	}
	return (xma-xmi)*(yma-ymi);
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n;
	for(Int i=0; i<n; ++i){
		int tx, ty;
		char c;
		cin >> tx >> ty >> c;
		if(c == 'R' || c == 'L'){
			y.push_back(P(ty, 0));
			x.push_back(P(tx, c=='R'?1:-1));
		}else{
			x.push_back(P(tx, 0));
			y.push_back(P(ty, c=='U'?1:-1));
		}
	}
	double tl = 0.0, tr = 1e10;
	for(int i=0; i<150; ++i){
		double t1 = (2 * tl + tr) / 3;
		double t2 = (tl + 2 * tr) / 3;
		double f = func(t1), g = func(t2);
		if(f < g){
			tr = t2;
		}else{
			tl = t1;
		}
	}
	cout << fixed << setprecision(14) << func(tl) << "\n";
}