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 syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<ll,pll> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-9;
const ll mod=1e9+7;
const int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};

ll n;
vl x,y,d,a;

ll f(ll m){
	ll mnx=inf,mxx=-inf,mny=inf,mxy=-inf;
	for(int i=0;i<n;i++){
		ll X=x[i]+dx[d[i]]*m,Y=y[i]+dy[d[i]]*m;
		mnx=min(mnx,X);
		mxx=max(mxx,X);
		mny=min(mny,Y);
		mxy=max(mxy,Y);
	}
	return (mxx-mnx)*(mxy-mny);
}

int main(){
	cin>>n;
	x=y=d=vl(n);
	for(int i=0;i<n;i++){
		char c;
		cin>>x[i]>>y[i]>>c;
		x[i]*=2;
		y[i]*=2;
		if(c=='R') d[i]=1;
		if(c=='D') d[i]=2;
		if(c=='L') d[i]=3;
	}
	a.push_back(inf);
	a.push_back(0);
	for(int i=0;i<4;i++){
		ll A=-inf,B=inf,C=inf;
		for(int j=0;j<n;j++){
			if(d[j]==0) C=min(C,x[j]);
			else if(d[j]==2) A=max(A,x[j]);
			else B=min(B,x[j]);
		}
		if(A<=B) a.push_back(B-A);
		if(B<=C) a.push_back(C-B);
		if(A<=C) a.push_back((C-A)/2);
		for(int j=0;j<n;j++){
			int X=y[j],Y=-x[j];
			x[j]=X;
			y[j]=Y;
			d[j]=(d[j]+1)%4;
		}
	}
	sort(a.begin(),a.end());
	ll N=a.size(),res=INF;
	for(int i=0;i<N-1;i++){
		ll l=a[i],r=a[i+1];
		while(r-l>5){
			ll m1=(r-l)/3+l,m2=(m1+r)/2;
			if(f(m1)<=f(m2)) r=m2;
			else l=m1;
		}
		for(int j=l;j<=r;j++) res=min(res,f(j));
	}
	cout<<syosu(11)<<(double)res/4<<endl;
}