#include<iostream>
#include<algorithm>
#include<vector>
#include<limits.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
const ll INF=LLONG_MAX/2;

int n;
vector<ll> x,y;
vector<int> dx,dy;

ll solve(ll t)
{
    ll xmin=INF;
    ll ymin=INF;
    ll xmax=-INF;
    ll ymax=-INF;
    rep(i,n)
    {
        ll x2=x[i]+dx[i]*t;
        ll y2=y[i]+dy[i]*t;
        xmin=min(xmin,x2);
        xmax=max(xmax,x2);
        ymin=min(ymin,y2);
        ymax=max(ymax,y2);
    }
    return (xmax-xmin)*(ymax-ymin);
}
int main()
{
    cin>>n;
    x.assign(n,0);
    y.assign(n,0);
    dx.assign(n,0);
    dy.assign(n,0);
    rep(i,n)
    {
        char d;
        cin>>x[i]>>y[i]>>d;
        x[i]*=2;
        y[i]*=2;
        if(d=='R')dx[i]++;
        else if(d=='L')dx[i]--;
        else if(d=='U')dy[i]++;
        else if(d=='D')dy[i]--;

    }
    ll left=0;
    ll right=400000000;
    while(right-left>3)
    {
        ll mid1=(left*2+right)/3;
        ll mid2=(left+right*2)/3;
        if(solve(mid1)<solve(mid2))right=mid2;
        else left=mid1;

    }
    ll ans=INF;
    for(ll i=left;i<=right;i++)ans=min(ans,solve(i));
    printf("%.10f\n",ans/4.0);
}
### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include<iostream>
#include<algorithm>
#include<vector>
#include<limits.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
const ll INF=LLONG_MAX/2;

int n;
vector<ll> x,y;
vector<int> dx,dy;

ll solve(ll t)
{
    ll xmin=INF;
    ll ymin=INF;
    ll xmax=-INF;
    ll ymax=-INF;
    rep(i,n)
    {
        ll x2=x[i]+dx[i]*t;
        ll y2=y[i]+dy[i]*t;
        xmin=min(xmin,x2);
        xmax=max(xmax,x2);
        ymin=min(ymin,y2);
        ymax=max(ymax,y2);
    }
    return (xmax-xmin)*(ymax-ymin);
}
int main()
{
    cin>>n;
    x.assign(n,0);
    y.assign(n,0);
    dx.assign(n,0);
    dy.assign(n,0);
    rep(i,n)
    {
        char d;
        cin>>x[i]>>y[i]>>d;
        x[i]*=2;
        y[i]*=2;
        if(d=='R')dx[i]++;
        else if(d=='L')dx[i]--;
        else if(d=='U')dy[i]++;
        else if(d=='D')dy[i]--;

    }
    ll left=0;
    ll right=400000000;
    while(right-left>3)
    {
        ll mid1=(left*2+right)/3;
        ll mid2=(left+right*2)/3;
        if(solve(mid1)<solve(mid2))right=mid2;
        else left=mid1;

    }
    ll ans=INF;
    for(ll i=left;i<=right;i++)ans=min(ans,solve(i));
    printf("%.10f\n",ans/4.0);
}
```