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;
typedef long long ll;
#define ft first
#define sd second
#define fr(i,n) for(int i=0;i<(n);++i)
#define Fr(i,n) for(int i=1;i<=(n);++i)
#define ifr(i,n) for(int i=(n)-1;i>=0;--i)
#define iFr(i,n) for(int i=(n);i>0;--i)

long double f(long double t,vector<vector<ll>>& k){
    long double tmp=0,r[4];
    fr(i,4) r[i]=max({k[i][0]+t,k[i][1]-t,(long double)k[i][2]});
    return (r[0]+r[2])*(r[1]+r[3]);
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    ll n,x,y;
    vector<vector<ll>> k(4,vector<ll>(3,-1e10));
    char d;
    cin>>n;
    fr(i,n){
        cin>>x>>y>>d;
        if(d=='R'){
            k[0][0]=max(k[0][0],x);
            k[1][2]=max(k[1][2],-y);
            k[2][1]=max(k[2][1],-x);
            k[3][2]=max(k[3][2],y);
        }
        if(d=='D'){
            k[0][2]=max(k[0][2],x);
            k[1][0]=max(k[1][0],-y);
            k[2][2]=max(k[2][2],-x);
            k[3][1]=max(k[3][1],y);
        }
        if(d=='L'){
            k[0][1]=max(k[0][1],x);
            k[1][2]=max(k[1][2],-y);
            k[2][0]=max(k[2][0],-x);
            k[3][2]=max(k[3][2],y);
        }
        if(d=='U'){
            k[0][2]=max(k[0][2],x);
            k[1][1]=max(k[1][1],-y);
            k[2][2]=max(k[2][2],-x);
            k[3][0]=max(k[3][0],y);
        }
    }
    long double ans=f(0,k);
    fr(i,4){
        if(k[i][2]-k[i][0]>0) ans=min(ans,f(k[i][2]-k[i][0],k));
        if(k[i][1]-k[i][2]>0) ans=min(ans,f(k[i][1]-k[i][2],k));
        if(k[i][1]-k[i][0]>0) ans=min(ans,f((long double)(k[i][1]-k[i][0])/2.0,k));
    }
    cout<<setprecision(20)<<ans<<endl;
}