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;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl





using ll = long long;


const int maxn = 1e6 + 5;
const ll inf = 1e12;


int n;
double x[maxn], y[maxn];
char d[maxn];


double sim(double t) {
    double xlo=inf,xhi=-inf,ylo=inf,yhi=-inf;
    for (int i=0; i<n; i++) {
	double X = x[i];
	double Y = y[i];
	if (d[i]=='U') Y += t;
	if (d[i]=='D') Y -= t;
	if (d[i]=='L') X -= t;
	if (d[i]=='R') X += t;
	xlo=min(xlo,X);
	xhi=max(xhi,X);
	ylo=min(ylo,Y);
	yhi=max(yhi,Y);
    }

    return (xhi-xlo)*(yhi-ylo);
}

const double eps = 1e-7;

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

    cin>>n;
    for (int i=0; i<n; i++) {
	cin>>x[i]>>y[i]>>d[i];
    }

    double lo = 0;
    double hi = 2e8;


    for (int iter=0; iter<500; iter++) {
	double mid1 = lo + (hi-lo)/3;
	double mid2 = hi - (hi-lo)/3;
	double res1 = sim(mid1);
	double res2 = sim(mid2);
	if (res1 < res2) {
	    hi = mid2;
	} else {
	    lo = mid1;	    
	}
    }

    cout<<fixed<<setprecision(12)<<sim(lo)<<endl;    
    return 0;
}