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:
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}


int nextInt() { int x; scanf("%d", &x); return x;}

const int MAX_N = 112345;
const ll INF = 1LL << 61;

int N;
ll X[MAX_N];
ll Y[MAX_N];
char D[MAX_N];

typedef pair<ll, ll> P;
P pos(int i, ll t) {
  switch (D[i]) {
    case 'R': return {X[i] + t, Y[i]};
    case 'L': return {X[i] - t, Y[i]};
    case 'U': return {X[i], Y[i] + t};
    case 'D': return {X[i], Y[i] - t};
  }
  return {0, 0};
}

ll f(ll t) {
  ll x_min = INF;
  ll x_max = -INF;
  ll y_min = INF;
  ll y_max = -INF;
  REP(i, N) {
    P p = pos(i, t);
    chmin(x_min, p.first);
    chmax(x_max, p.first);
    chmin(y_min, p.second);
    chmax(y_max, p.second);
  }
  return (x_max - x_min) * (y_max - y_min);
}

int main2() {
  N = nextInt();
  REP(i, N) {
    X[i] = nextInt() * 2;
    Y[i] = nextInt() * 2;
    cin >> D[i];
  }

  ll lo = 0;
  ll hi = 1e9;
  while (hi - lo >= 3) {
    ll l = (lo * 2 + hi) / 3;
    ll r = (lo + hi * 2) / 3;
    if (f(l) > f(r)) lo = l; else hi = r; // min: > , max: <
  }
  ll minVal = f(lo);
  for (ll i = lo + 1; i <= hi; i++)
    minVal = min(minVal, f(i));
  printf("%.10f\n", minVal / 4.0);
  return 0;
}

int main() {

#ifdef LOCAL
  for (;!cin.eof();cin>>ws)
#endif
    main2();
  return 0;
}