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 N = 1e5 + 7;
const double inf = 1e18;

struct Point {
	double x, y;
	int k1, k2;
} p[N];

int n;

double check(double mid) {
	double xmx = -inf, xmn = inf, ymx = -inf, ymn = inf;
	for (int i = 0; i < n; i++) {
		xmx = max(xmx, p[i].x + p[i].k1 * mid);
		xmn = min(xmn, p[i].x + p[i].k1 * mid);
		ymx = max(ymx, p[i].y + p[i].k2 * mid);
		ymn = min(ymn, p[i].y + p[i].k2 * mid);
	}
	return (xmx - xmn) * (ymx - ymn);
}

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		char s[10];
		double x, y;
		scanf("%lf%lf%s", &x, &y, s);
		p[i].x = x, p[i].y = y;
		if (s[0] == 'R') p[i].k1 = 1;
		else if (s[0] == 'L') p[i].k1 = -1;
		else if (s[0] == 'U') p[i].k2 = 1;
		else p[i].k2 = -1;
	}
	double l = 0, r = 1e9;
	double ans = inf;
	for (int i = 0; i < 200; i++) {
		double ll = l + (r - l) / 3, rr = r - (r - l) / 3;
		double temp1 = check(ll), temp2 = check(rr);
		ans = min(ans, min(temp1, temp2));
		if (temp1 > temp2) l = ll;
		else r = rr;
		
	}
	printf("%.15f\n", ans);
	return 0;
}