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 <iostream>
#include <string>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <iomanip>


using namespace std;

#define REP(i, n) for(ll i = 0;i < n;i++)
#define REPR(i, n) for(ll i = n;i >= 0;i--)
#define FOR(i, m, n) for(ll i = m;i < n;i++)
#define FORR(i, m, n) for(ll i = m;i >= n;i--)
#define REPO(i, n) for(ll i = 1;i <= n;i++)
#define ll long long
#define INF (ll)1 << 60
#define MINF (-1 * INF)
#define ALL(n) n.begin(),n.end()
#define MOD (ll)1000000007
#define P pair<ll, ll>

ll n;
double x[110000], y[110000];
char d[110000];
const double EPS = 1e-10;

double f(double a) {
	double xmx = 1e8 * 3 * -1, xmi = 1e8 * 3, ymx = 1e8 * 3 * -1, ymi = 1e8 * 3;
	REP(i, n) {
		double nowx = x[i], nowy = y[i];
		if (d[i] == 'R') nowx += a;
		if (d[i] == 'L') nowx -= a;
		if (d[i] == 'U') nowy += a;
		if (d[i] == 'D') nowy -= a;
		xmx = max(xmx, nowx);
		xmi = min(xmi, nowx);
		ymx = max(ymx, nowy);
		ymi = min(ymi, nowy);
		//cout << nowx << " " << nowy << endl;
	}
	//cout << xmx << " " << xmi << " " << ymx << " " << ymi << endl;
	return (xmx - xmi) * (ymx - ymi);
}

int main() {
	cin >> n;
	REP(i, n) {
		cin >> x[i] >> y[i] >> d[i];
	}
	double low = 0.0, high = 1e8 * 3;
	REP(i,300) {
		double width = high - low, ml = low + width / 3.0, mr = low + 2.0 * width / 3.0;
		if (f(ml) <= f(mr)) {
			high = mr;
		}
		else {
			low = ml;
		}
	}
	//cout << (high + low) * 0.5 << endl;
	cout << setprecision(10) << fixed << f((high + low) * 0.5) << endl;
}