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

Solution (in CPP):

Solution:
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll,ll> pii;
typedef long double ld;

#define fi first
#define se second
#define mp make_pair
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ld EPS = 6e-11;
const int N = (int)1e5 + 10;
ld x[N];
ld y[N];
char dir[N];
int n;

ld fx(ld p){
  ld lx=(ld)1e9,rx=-(ld)1e9;
  ld cx;
  for(int i = 0 ; i < n; i ++ ){
    cx = x[i];
    if(dir[i]=='L')cx-=p;
    if(dir[i]=='R')cx+=p;
    lx=min(lx,cx);
    rx=max(rx,cx);
  }
  return (rx - lx);
}

ld fy(ld p){
  ld ly=(ld)1e9,ry=-(ld)1e9;
  ld cy;
  for(int i = 0 ; i < n; i ++ ){
    cy = y[i];
    if(dir[i]=='D')cy-=p;
    if(dir[i]=='U')cy+=p;
    ly=min(ly,cy);
    ry=max(ry,cy);
  }
  return (ry-ly);
}


int main(){
  fastIO;
  cin >> n;
  for(int i = 0; i < n; i ++ ){
    cin >> x[i] >> y[i] >> dir[i];
  }
  ld ans = fx(0)*fy(0);
  ld li = 0, ri = (ld)4e8;
  ld m1, m2;
  ld f1, f2;
  for(int it = 0; it < 300; it ++ ){
    m1 = (li + li + ri) / 3.0;
    m2 = (li + ri + ri) / 3.0;
    f1 = fx(m1)*fy(m1);
    f2 = fx(m2)*fy(m2);
    ans = min(ans, f1);
    ans = min(ans, f2);
    if(fx(m1)*fy(m1)>fx(m2)*fy(m2)){
      li = m1;
    }
    else{
      ri = m2;
    }
  }
  cout << fixed << setprecision(12) << ans << "\n";
  return 0;
}