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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma gcc optimize("Ofast")

#include "bits/stdc++.h"
using namespace std;
typedef int64_t lld;
typedef pair<int,int> pii;
typedef pair<lld,lld> pll;
typedef pair<int,pll> pip;
typedef pair<pll,int> ppi;
typedef pair<lld,pll> plp;
typedef pair<pll,lld> ppl;
typedef pair<pll,pll> ppp;
template<typename T>
using maxHeap = priority_queue<T,vector<T>,less<T>>;
template<typename T>
using minHeap = priority_queue<T,vector<T>,greater<T>>;
#define ff first
#define ss second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define endl '\n'
#define jizz cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
inline void input(int &_x) {
    _x = 0;
    int _tmp = 1; char _tc = getchar();
    while((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
    if(_tc == '-') _tc = getchar(), _tmp = -1;
    while(_tc >= '0' && _tc <= '9') _x = _x*10+(_tc-48), _tc = getchar();
    _x *= _tmp;
}
inline void output(int _x) {
    char _buff[20]; int _f = 0;
    if(_x == 0)putchar('0');
    while(_x > 0)
    {
        _buff[_f++] = _x%10+'0';
        _x /= 10;
    }
    for(_f-=1; _f >= 0; _f--)
        putchar(_buff[_f]);
    putchar('\n');
}
template<typename Iter>
ostream& _out(ostream &s, Iter b, Iter e) {
    s<<"[";
    for ( auto it=b; it!=e; it++ ) s<<(it==b?"":" ")<<*it;
    s<<"]";
    return s;
}
template<class T1,class T2>
ostream& operator<<(ostream& out, pair<T1,T2> p) {
    return out << '(' << p.first << ", " << p.second << ')';
}
template<typename T>
ostream& operator <<( ostream &s, const vector<T> &c ) {
     return _out(s,c.begin(),c.end());
 }
#ifdef erd1
    #define pprint(x) cerr<<__PRETTY_FUNCTION__<<":"<<__LINE__<<" - "<<(#x)<<"="<<(x)<<endl
#else
    #define pprint(x)
#endif

// code starts here
void upd(pii& p, int x){
    p.ff = min(p.ff, x);
    p.ss = max(p.ss, x);
}
pii v, h, u, d, l, r;
lld calc(int t){
    if(t<0)return LLONG_MAX;
    //cout << t << endl;
    int L = min({r.ff+t, h.ff, l.ff-t}),
        R = max({r.ss+t, h.ss, l.ss-t}),
        U = max({u.ss+t, v.ss, d.ss-t}),
        D = min({u.ff+t, v.ff, d.ff-t});
    return (lld)(R-L)*(U-D);
}
signed main(){
    int n;
    v = h = u = d = l = r = {5e8, -5e8};
    cin >> n;
    for(int i = 0; i < n; i++){
        int a, b; char c;
        cin >> a >> b >> c;a*=2;b*=2;
        switch (c) {
            case 'L': upd(v, b);upd(l, a);break;
            case 'R': upd(v, b);upd(r, a);break;
            case 'U': upd(h, a);upd(u, b);break;
            case 'D': upd(h, a);upd(d, b);break;
        }
    }
    //cout <<v<<h<<u<<d<<l<<r<<endl;
    cout << min({ calc(d.ss-v.ss), calc(v.ss-u.ss), calc((d.ss-u.ss)/2),
                  calc(d.ff-v.ff), calc(v.ff-u.ff), calc((d.ff-u.ff)/2),
                  calc(l.ss-h.ss), calc(h.ss-r.ss), calc((l.ss-r.ss)/2),
                  calc(l.ff-h.ff), calc(h.ff-r.ff), calc((l.ff-r.ff)/2),
                  calc(0) })/4.0 << endl;
}