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;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

int main(){
  int n;
  cin >> n;
  vector<pair<int,int>> p(n),d(n);
  map<char,pair<int,int>> mp;
  mp['R']={1,0};
  mp['L']={-1,0};
  mp['U']={0,1};
  mp['D']={0,-1};
  char c;
  rep(i,n){
    cin >> p[i].first >> p[i].second;
    cin >> c;
    d[i]=mp[c];
  }
  double l=0,r=200000000,mid1,mid2;
  double ans=LINF;
  rep(i,100){
    mid1=l+(r-l)/3;
    mid2=l+2*(r-l)/3;
    double c1,c2;
    double lx=IINF,rx=-IINF,ly=IINF,ry=-IINF;
    rep(i,n){
      double x = p[i].first+d[i].first*mid1;
      double y = p[i].second+d[i].second*mid1;
      lx=min(lx,x);
      rx=max(rx,x);
      ly=min(ly,y);
      ry=max(ry,y);
    }
    c1=(rx-lx)*(ry-ly);
    lx=IINF,rx=-IINF,ly=IINF,ry=-IINF;
    rep(i,n){
      double x = p[i].first+d[i].first*mid2;
      double y = p[i].second+d[i].second*mid2;
      lx=min(lx,x);
      rx=max(rx,x);
      ly=min(ly,y);
      ry=max(ry,y);
    }
    c2=(rx-lx)*(ry-ly);
    if(c1<c2){
      ans=min(ans,c1);
      r=mid2;
    }else{
      ans=min(ans,c2);
      l=mid1;
    }
  }
  cout << fixed << setprecision(10) << ans << endl;
  return 0;
}