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 maxn = 100100;
int x[maxn], y[maxn]; char d[maxn];
int main(){
	int n; cin >> n;
	for(int i = 0; i < n; i++) cin >> x[i] >> y[i] >> d[i];
	auto apply = [&](double t, int ind){
		pair<double, double> ans;
		ans.first = x[ind];
		ans.second = y[ind];
		if(d[ind] == 'L') ans.first -= t;
		if(d[ind] == 'R') ans.first += t;
		if(d[ind] == 'U') ans.second += t;
		if(d[ind] == 'D') ans.second -= t;
		return ans;
	};
	auto fdx = [&](double t){
		double maxx, minx, maxy, miny;
		maxx = minx = apply(t, 0).first;
		maxy = miny = apply(t, 0).second;
		for(int i = 1; i < n; i++){
			auto r = apply(t, i);
			maxx = max(maxx, r.first);
			minx = min(minx, r.first);
			maxy = max(maxy, r.second);
			miny = min(miny, r.second);
		}
		return (maxx - minx) * (maxy - miny);
	};
	double lo = 0, hi = 1e9;
	for(int i = 0; i < 200; i++){
		double p0 = lo + (hi - lo) / 3;
		double p1 = hi - (hi - lo) / 3;
		if(fdx(p0) < fdx(p1)) hi = p1;
		else lo = p0;
	}
	cout << fixed << setprecision(12) << fdx(lo) << endl;
	return 0;
}