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<algorithm>
#include<vector>
#include<queue>
#include<unordered_set>
#include<unordered_map>
#include<iomanip>
using namespace std;
typedef long long ll;
#define mod 1000000007
#define mad(a,b) a=(a+b)%mod;
#define chmin(a,b) a=min(a,b)
#define chmax(a,b) a=max(a,b)
typedef long double ld;
ld mi[4],ma[4];
ld mixx,maxx,miyy,mayy;
ld solve(ld k){
    ld mx=min(mixx,min(mi[0]+k,mi[1]-k));
    ld ex=max(maxx,max(ma[0]+k,ma[1]-k));
    ld my=min(miyy,min(mi[2]+k,mi[3]-k));
    ld ey=max(mayy,max(ma[2]+k,ma[3]-k));
    return (ex-mx)*(ey-my);
}
int main(){
  ll n;cin>>n;
  for(int i=0;i<4;i++){
      mi[i]=1e17;
      ma[i]=-1e17;
  }
  maxx=mayy=-1e17;
  miyy=mixx=1e17;
  for(int i=0;i<n;i++){char z;ll d;
      ld x,y;cin>>x>>y>>z;
      if(z=='R')d=0;
      if(z=='L')d=1;
      if(z=='U')d=2;
      if(z=='D')d=3;
      if(d<2){
      chmin(mi[d],x);
      chmax(ma[d],x);
      chmin(miyy,y);
      chmax(mayy,y);
      }
      else{
	  chmin(mi[d],y);
	chmax(ma[d],y);
	chmin(mixx,x);
	chmax(maxx,x);
      }
  }
  ld l=0,r=1e10,p,q;
  for(int i=0;i<10000;i++){
      q=(l+r*2)/3.0;
      p=(l*2+r)/3.0;
      if(solve(p)>solve(q))l=p;
      else r=q;
  }  //cout<<r<<endl;
  ld res=solve(r);
  cout<<fixed<<setprecision(10)<<res<<endl;
  return 0;
}