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 prArr(Arr,n) for(int _=0;_<n;_++)   cout<<(Arr)[_]<<" ";
#define fi first
#define s second
using namespace std;
const int INF=(int)0x3f3f3f3f;
const int MOD=(int)1e9+7;
const double eps=1e-9;
int n;
vector<pair<pair<int,int>,pair<int,int>>> v;
double f(double t)
{
    double inf=(double)INF*INF;
    double minX=inf,minY=inf,maxX=-inf,maxY=-inf;
    for(int i=0;i<n;i++){
        double x=v[i].fi.fi+t*v[i].s.fi,y=v[i].fi.s+t*v[i].s.s;
        minX=min(minX,x),maxX=max(maxX,x);
        minY=min(minY,y),maxY=max(maxY,y);
    }
    //cout<<minX<<" "<<maxX<<" "<<minY<<" "<<maxY<<"\n";
    return (maxX-minX)*(maxY-minY);
}
double ternary_search(double l, double r) {

    for(int i=0;i<1000;i++) {
        double m1 = l + (r - l) / 3;
        double m2 = r - (r - l) / 3;
        double f1 = f(m1);      //evaluates the function at m1
        double f2 = f(m2);      //evaluates the function at m2
        //cout<<m1<<" "<<f1<<" "<<m2<<" "<<f2<<"\n";
        if (f1 < f2)
            r = m2;
        else
            l = m1;
    }
    return f(l);                    //return the maximum of f(x) in [l, r]
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    char c;
    cin>>n;
    v.resize(n);
    for(int i=0;i<n&&cin>>v[i].fi.fi>>v[i].fi.s>>c;i++)
    {
        if(c=='U')v[i].s={0,1};
        if(c=='D')v[i].s={0,-1};
        if(c=='R')v[i].s={1,0};
        if(c=='L')v[i].s={-1,0};
    }
    cout<<setprecision(10)<<fixed<<ternary_search(0,1e9);
    return 0;
}