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 <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
vector<int> X;
vector<int> Y;
vector<int> DX;
vector<int> DY;
int N;
double solve(double a)
{

	double mxx = -1e18;
	double mnx = 1e18;
	double mxy = -1e18;
	double mny = 1e18;


	for (int i = 0; i < N; i++)
    {
		double x = X[i] + a * DX[i];
		double y = Y[i] + a * DY[i];

		mxx = max(mxx, x);
		mnx = min(mnx, x);

		mxy = max(mxy, y);
		mny = min(mny, y);
	}

	return (mxx - mnx) * (mxy - mny);
}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> N;
	X.resize(N);
	Y.resize(N);
	DX.resize(N, 0);
	DY.resize(N, 0);
	char dd;
	for (int i = 0; i < N; i++)
    {
		cin >> X[i] >> Y[i] >> dd;

		if (dd == 'U')
        {
			DY[i] = 1;
		}
		else if (dd == 'D')
		{
			DY[i] = -1;
		}
		else if (dd == 'R')
		{
			DX[i] = 1;
		}
		else if (dd == 'L')
        {
			DX[i] = -1;
		}
	}
	double res = 1e18;

	double u = 1e9;
	double d = 0;

	for (int t = 0; t < 200; t++)
    {

		double uu = (u * 2 + d) / 3;
		double dd = (u + d * 2) / 3;

		double uv = solve(uu);
		double dv = solve(dd);
		res = min(res, uv);
		res = min(res, dv);
		if (dv < uv)
        {
			u = uu;
		}
		else
		{
			d = dd;
		}

	}


	cout << fixed << setprecision(15) << res << endl;
}