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 mk make_pair
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef long double ld;
ll x[500010], y[500010];
char d[500010][10];
vector <ll> v1, v2;
ld val(ld xx){
    ld mx1=-1e9, mn1=1e9, mx2=-1e9, mn2=1e9, tmp;
    for(int i=0; i<v1.size(); ++i){
        if(d[v1[i]][0]=='L'){
            tmp=x[v1[i]]-xx;
        }
        else{
            tmp=x[v1[i]]+xx;
        }
        mx1=max(mx1, tmp);
        mn1=min(mn1, tmp);
        mx2=max(mx2, (ld)y[v1[i]]);
        mn2=min(mn2, (ld)y[v1[i]]);
    }

    for(int i=0; i<v2.size(); ++i){
        if(d[v2[i]][0]=='D'){
            tmp=y[v2[i]]-xx;
        }
        else{
            tmp=y[v2[i]]+xx;
        }
        mx2=max(mx2, tmp);
        mn2=min(mn2, tmp);
        mx1=max(mx1, (ld)x[v2[i]]);
        mn1=min(mn1, (ld)x[v2[i]]);
    }
    return (mx1-mn1)*(mx2-mn2);
}
int main(){
    ll n;
    while(cin>>n){
        ld ans=0;
        for(ll i=0; i<n; ++i){
            scanf("%lld%lld%s",x+i,y+i,d+i);
            if(d[i][0]=='L'||d[i][0]=='R'){
                v1.push_back(i);
            }
            else{
                v2.push_back(i);
            }
        }
        ld st=0, nd=3e8, t1, t2, tmp1, tmp2;
        int t=200;
        while(t--){
            t1=st+(nd-st)/3.0;
            t2=nd-(nd-st)/3.0;
            tmp1=val(t1);
            tmp2=val(t2);
            if(tmp1>tmp2){
                st=t1;
            }
            else{
                nd=t2;
            }
        }
        ans=val(nd);
        cout<<fixed<<setprecision(12)<<ans<<endl;
    }
  return 0;
}