#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7,N=1e5+5;
int mul(int a,int b){return (long long)a*b;}
typedef long long ll;
int n,x[N],y[N],d[N];
const double inf=1e20;
double getans(double mid)
{
    double mnx=inf,mxx=-inf,mny=inf,mxy=-inf;
    for(int i=1;i<=n;i++)
    {
        double xx=x[i],yy=y[i];
        if(d[i]==1) xx-=mid;
        if(d[i]==2) xx+=mid;
        if(d[i]==3) yy+=mid;
        if(d[i]==4) yy-=mid;
        mnx=min(mnx,xx);
        mxx=max(mxx,xx);
        mny=min(mny,yy);
        mxy=max(mxy,yy);
    }
    return (mxx-mnx)*(mxy-mny);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        char s[5];
        scanf("%d%d%s",&x[i],&y[i],s);
        if(s[0]=='L') d[i]=1;
        else if(s[0]=='R') d[i]=2;
        else if(s[0]=='U') d[i]=3;
        else d[i]=4;
    }
    double l=0,r=1e10;
    for(int i=1;i<=1000;i++)
    {
        double mid=l+(r-l)/3,midmid=l+(r-l)/3*2;
        if(getans(mid)<getans(midmid)) r=midmid;
        else l=mid;
    }
    printf("%f\n",getans(l));
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7,N=1e5+5;
int mul(int a,int b){return (long long)a*b;}
typedef long long ll;
int n,x[N],y[N],d[N];
const double inf=1e20;
double getans(double mid)
{
    double mnx=inf,mxx=-inf,mny=inf,mxy=-inf;
    for(int i=1;i<=n;i++)
    {
        double xx=x[i],yy=y[i];
        if(d[i]==1) xx-=mid;
        if(d[i]==2) xx+=mid;
        if(d[i]==3) yy+=mid;
        if(d[i]==4) yy-=mid;
        mnx=min(mnx,xx);
        mxx=max(mxx,xx);
        mny=min(mny,yy);
        mxy=max(mxy,yy);
    }
    return (mxx-mnx)*(mxy-mny);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        char s[5];
        scanf("%d%d%s",&x[i],&y[i],s);
        if(s[0]=='L') d[i]=1;
        else if(s[0]=='R') d[i]=2;
        else if(s[0]=='U') d[i]=3;
        else d[i]=4;
    }
    double l=0,r=1e10;
    for(int i=1;i<=1000;i++)
    {
        double mid=l+(r-l)/3,midmid=l+(r-l)/3*2;
        if(getans(mid)<getans(midmid)) r=midmid;
        else l=mid;
    }
    printf("%f\n",getans(l));
}

```