#include <bits/stdc++.h>
#define dbug(x) cout<<#x<<"="<<x<<endl
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const double INF=(1e20);
const int maxn=(1e5)+10;
int n; char s[10];
int id[300],a[maxn];
int x[maxn],y[maxn];
double xmx,xmn,ymx,ymn;
int fx[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
double solve(double mid) {
	xmx=ymx=-INF; xmn=ymn=INF;
	double A,B;
	for (int i=1;i<=n;i++) {
		A=x[i],B=y[i];
		if (a[i]==0) A+=mid;
		else if (a[i]==1) A-=mid;
		else if (a[i]==2) B+=mid;
		else B-=mid;
		xmx=max(xmx,A);
		xmn=min(xmn,A);
		ymx=max(ymx,B);
		ymn=min(ymn,B);
	//	printf("%.2lf %.2lf\n",A,B);
	}
	return (xmx-xmn)*(ymx-ymn);
}
int main() {
	//freopen("1.txt","r",stdin);
	read(n);
	id['R']=0; id['L']=1;
	id['U']=2; id['D']=3;
	for (int i=1;i<=n;i++) {
		read(x[i]),read(y[i]);
		scanf("%s",s+1);
		a[i]=id[s[1]];
	}
	double l=0,r=(1e10),m1,m2,len;
	for (int i=1;i<=1000;i++) {
		len=(r-l)/3;
		m1=l+len; m2=l+2*len;
		if (solve(m1)<solve(m2)) r=m2;
		else l=m1;
	}
	printf("%f\n",solve(l));
	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 <bits/stdc++.h>
#define dbug(x) cout<<#x<<"="<<x<<endl
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const double INF=(1e20);
const int maxn=(1e5)+10;
int n; char s[10];
int id[300],a[maxn];
int x[maxn],y[maxn];
double xmx,xmn,ymx,ymn;
int fx[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
double solve(double mid) {
	xmx=ymx=-INF; xmn=ymn=INF;
	double A,B;
	for (int i=1;i<=n;i++) {
		A=x[i],B=y[i];
		if (a[i]==0) A+=mid;
		else if (a[i]==1) A-=mid;
		else if (a[i]==2) B+=mid;
		else B-=mid;
		xmx=max(xmx,A);
		xmn=min(xmn,A);
		ymx=max(ymx,B);
		ymn=min(ymn,B);
	//	printf("%.2lf %.2lf\n",A,B);
	}
	return (xmx-xmn)*(ymx-ymn);
}
int main() {
	//freopen("1.txt","r",stdin);
	read(n);
	id['R']=0; id['L']=1;
	id['U']=2; id['D']=3;
	for (int i=1;i<=n;i++) {
		read(x[i]),read(y[i]);
		scanf("%s",s+1);
		a[i]=id[s[1]];
	}
	double l=0,r=(1e10),m1,m2,len;
	for (int i=1;i<=1000;i++) {
		len=(r-l)/3;
		m1=l+len; m2=l+2*len;
		if (solve(m1)<solve(m2)) r=m2;
		else l=m1;
	}
	printf("%f\n",solve(l));
	return 0;
}
```