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)

#include <bits/stdc++.h>
#include <iomanip>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define fi first
#define se second
#define mp make_pair
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

ll N,M,H,W,K,Q,A,B,L,R;
string S,T;
const ll MOD = (1e+9) + 7;
const ll INF = 1LL << 60;
typedef pair<ll,ll> P;

typedef vector<P> vp;
typedef vector<vp> matP;

const ll MAX_N = 100010;
vec x(MAX_N),y(MAX_N);
vector<char> d(MAX_N);
map<char,P> dir = {mp('L',mp(-1,0)),mp('R',mp(1,0)),mp('D',mp(0,-1)),mp('U',mp(0,1))};
ll simu(ll t_2,bool X, bool mini){ //tは二倍で受け取る
    ll temp = mini ? 1e+10 : -1e+10;
    rep(i,N){
        if(X){
            temp = mini ? min(temp,x[i] + dir[d[i]].fi*t_2) : max(temp,x[i] + dir[d[i]].fi*t_2);
        }else{
            temp = mini ? min(temp,y[i] + dir[d[i]].se*t_2) : max(temp,y[i] + dir[d[i]].se*t_2);
        }
    }
    return temp;
}

int main() {
    cin>>N;
    rep(i,N) cin>>x[i]>>y[i]>>d[i];
    rep(i,N){
        x[i] *= 2;
        y[i] *= 2;
    }
    set<ll> time;//2倍で保存
    time.insert(0LL);
    rep(i,2) {
        rep(j,2) {
            ll lb = -5, ub = 4e+8; // ub,lbはtimeの2倍
            while (ub - lb > 1) {
                ll cen = (ub + lb) / 2;
                (simu(cen, j, i) > simu(cen + 1, j, i) ? lb : ub) = cen;
            }
            rep(i,7)time.insert(max(0LL,ub+i-3));
        }
    }
    rep(i,2) {
        rep(j,2) {
            ll lb = -5, ub = 4e+8; // ub,lbはtimeの2倍
            while (ub - lb > 1) {
                ll cen = (ub + lb) / 2;
                (simu(cen, j, i) < simu(cen + 1, j, i) ? ub : lb) = cen;
            }
            rep(i,7)time.insert(max(0LL,ub+i-3));
        }
    }
    long double ans = 1e+18;
    for(ll T2 : time){
        //cout<<T<<endl;
        ll MAX, MIX,MAY,MIY;
        MAX = MAY = -2e+8;
        MIX = MIY = 2e+8;
        rep(i,N){
            MAX = max(MAX, x[i] + T2*dir[d[i]].fi);
            MIX = min(MIX, x[i] + T2*dir[d[i]].fi);
            MAY = max(MAY, y[i] + T2*dir[d[i]].se);
            MIY = min(MIY, y[i] + T2*dir[d[i]].se);
            //if(T2 == 2e+8) cout<<i<<' '<<MIX<<' '<<x[i]<<endl;
        }
        ll xr = MAX - MIX, yr = MAY - MIY;
        long double temp = xr*yr;
        temp /= 4.0;
        if(temp >= 0) ans = min(ans,temp);
        if(xr == 0 || yr == 0) {
            ans = 0;
            break;
        }
    }
    cout<<fixed<<setprecision(10)<<ans<<endl;
}