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>
#define ll long long
#define ld long double

using namespace std;

const ld EPS = 0.0000000001;
const ld INF = 1e8 + 10;

struct Data {
	ld x, y;
	int dir;	
};

const ld xplus[] = {1, -1, 0, 0};
const ld yplus[] = {0, 0, 1, -1};

ld check(ld t, const vector<Data> &v, int n) {
	ld xmax = -INF, xmin = INF, ymax = -INF, ymin = INF;	
	for(int i = 1; i <= n; i ++) {
		xmax = max(xmax, v[i].x + xplus[v[i].dir] * t);
		xmin = min(xmin, v[i].x + xplus[v[i].dir] * t);

		ymax = max(ymax, v[i].y + yplus[v[i].dir] * t);
		ymin = min(ymin, v[i].y + yplus[v[i].dir] * t);
	}

	ld ans = (xmax - xmin) * (ymax - ymin);
	return ans;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n;
	cin >> n;
	vector<Data> v(n + 1);
	for(int i = 1; i <= n; i ++) {
		cin >> v[i].x >> v[i].y;
		char c;
		cin >> c;
		if(c == 'R')
			v[i].dir = 0;
		if(c == 'L')
			v[i].dir = 1;
		if(c == 'U')
			v[i].dir = 2;
		if(c == 'D')
			v[i].dir = 3;
	}

	ld left = 0, right = 1e8;
	for(int step = 1; step < 200; step ++) {
		ld a = (left * 2 + right) / 3;
		ld b = (left + right * 2) / 3;
		if(check(a, v, n) >=  check(b, v, n))
			left = a;
		else
			right = b;
	//	cout << setprecision(10) << fixed << a  << " " << b << endl;  
	}
	//cout << setprecision(10) << fixed << check(left, v, n) << " " << check(0, v, n) << endl;

	cout << setprecision(12) << fixed << check(left, v, n);	

	return 0;
}