#include<cstdio>
#include<algorithm>
#define maxn 100005
#define ll long long
using namespace std;
const double INF = 1e18;
int n;
double ans = INF;
ll x[maxn],y[maxn];
int d[maxn];
int dx[4] = {-1,1,0,0};
int dy[4] = {0,0,-1,1};

double solve(double t)
{
    double maxx = -INF,maxy = -INF;
    double minx = INF,miny = INF;
    double xx,yy;
    for (int i=1;i<=n;i++)
    {
        xx = x[i] + t*dx[d[i]];
        yy = y[i] + t*dy[d[i]];
        maxx = max(maxx,xx);
        minx = min(minx,xx);
        maxy = max(maxy,yy);
        miny = min(miny,yy);
    }
    return (maxx-minx)*(maxy-miny);
}

int main()
{
    scanf("%d",&n);
    char temp;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld %lld %c",&x[i],&y[i],&temp);
        if (temp=='L') d[i] = 0;
        else if(temp=='R') d[i] = 1;
        else if(temp=='D') d[i] = 2;
        else if(temp=='U') d[i] = 3;
    }
    double u=1e9;
    double d=0;
    for (int i=1;i<=200;i++)
    {
        double uu = (u*2+d)/3;
        double dd = (u+d*2)/3;
        double uv = solve(uu);
        double dv = solve(dd);
        ans = min(ans,uv);
        ans = min(ans,dv);
        if(uv>dv) u=uu;
        else d=dd;
        //printf("%f\n",ans);
    }
    printf("%f",ans);
    return 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<cstdio>
#include<algorithm>
#define maxn 100005
#define ll long long
using namespace std;
const double INF = 1e18;
int n;
double ans = INF;
ll x[maxn],y[maxn];
int d[maxn];
int dx[4] = {-1,1,0,0};
int dy[4] = {0,0,-1,1};

double solve(double t)
{
    double maxx = -INF,maxy = -INF;
    double minx = INF,miny = INF;
    double xx,yy;
    for (int i=1;i<=n;i++)
    {
        xx = x[i] + t*dx[d[i]];
        yy = y[i] + t*dy[d[i]];
        maxx = max(maxx,xx);
        minx = min(minx,xx);
        maxy = max(maxy,yy);
        miny = min(miny,yy);
    }
    return (maxx-minx)*(maxy-miny);
}

int main()
{
    scanf("%d",&n);
    char temp;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld %lld %c",&x[i],&y[i],&temp);
        if (temp=='L') d[i] = 0;
        else if(temp=='R') d[i] = 1;
        else if(temp=='D') d[i] = 2;
        else if(temp=='U') d[i] = 3;
    }
    double u=1e9;
    double d=0;
    for (int i=1;i<=200;i++)
    {
        double uu = (u*2+d)/3;
        double dd = (u+d*2)/3;
        double uv = solve(uu);
        double dv = solve(dd);
        ans = min(ans,uv);
        ans = min(ans,dv);
        if(uv>dv) u=uu;
        else d=dd;
        //printf("%f\n",ans);
    }
    printf("%f",ans);
    return 0;
}

```