Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

2
0 3 D
3 0 L

Sample Output 1

0

After three seconds, the two points will meet at the origin. The value in question will be 0 at that moment.

Sample Input 2

5
-7 -10 U
7 -6 U
-8 7 D
-3 3 D
0 -6 R

Sample Output 2

97.5

The answer may not be an integer.

Sample Input 3

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

Sample Output 3

273

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)(((n)>>(m))&1)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p)pom(a,p-2,p)

ll x[100010],y[100010];
char d[100010];

ll t[100],tcnt;
ll f[4][3];


int main(){
	ll n;
	scanf("%lld",&n);
	rep(i,0,n)scanf("%lld%lld %c",y+i,x+i,d+i);
	
	rep(i,0,4)rep(j,0,3)f[i][j]=-INF;

	rep(i,0,n){
		if(d[i]=='R')f[0][0]=max(f[0][0],y[i]);
		else if(d[i]=='L')f[0][1]=max(f[0][1],y[i]);
		else f[0][2]=max(f[0][2],y[i]);
		
		if(d[i]=='L')f[1][0]=max(f[1][0],-y[i]);
		else if(d[i]=='R')f[1][1]=max(f[1][1],-y[i]);
		else f[1][2]=max(f[1][2],-y[i]);

		if(d[i]=='U')f[2][0]=max(f[2][0],x[i]);
		else if(d[i]=='D')f[2][1]=max(f[2][1],x[i]);
		else f[2][2]=max(f[2][2],x[i]);
		
		if(d[i]=='D')f[3][0]=max(f[3][0],-x[i]);
		else if(d[i]=='U')f[3][1]=max(f[3][1],-x[i]);
		else f[3][2]=max(f[3][2],-x[i]);
	}
	
	double ans=INF;
	rep(t,0,400000010){
		double time=t/2.0;
		double a[4];
		rep(k,0,4)a[k]=max(f[k][0]+time,max(f[k][1]-time,f[k][2]));
		ans=min(ans,(a[0]+a[1])*(a[2]+a[3]));
	}
	printf("%.9f",ans);	
}