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<ext/pb_ds/assoc_container.hpp>
    #include<ext/pb_ds/tree_policy.hpp>
    #define pb push_back
    #define f first
    #define s second
    #define lint long long
    #define mp make_pair
    #define pii pair<int, int>
    #define pll pair<lint, lint>
    #define ld long double
    #define inlld(x) scanf("%lld", &x)
    #define ind(x) scanf("%d", &x)
    #define vi vector<int>
     
    const int N=1e5+5;
    const int MOD=1e9+7;
    const lint INF=1e18;
     
    using namespace std;
    using namespace __gnu_pbds;
     
    template <typename T> 
    using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
     
    #define pd pair < long double, long double > 
    #define PR 0.0000000001 
     
    pd points[N] ; 
    int ind ; 
    char dir[N] ; int n ; 
     
    pd giveNew(int i, ld t)
    {
        pd cur = points[i] ; 
        if(dir[i] == 'R') cur.first += t ;
        else if(dir[i] == 'L') cur.first -= t ;
        else if(dir[i] == 'U') cur.second += t ;
        else cur.second -= t ;
        return cur ; 
     
    }
     
    ld maxArea(ld t)
    {
        ld maxX = -1e9, minX = 1e9, maxY = -1e9, minY = 1e9 ;
        for(int i = 0; i < n; i++)
        {
            pd shift = giveNew(i, t);
            maxX = max(maxX, shift.first);
            minX = min(minX, shift.first);
            minY = min(minY, shift.second);
            maxY = max(maxY, shift.second);
        }
        return (maxX - minX)*(maxY - minY);
    }
     
    signed main()
    {
        cout<<fixed<<setprecision(10); 
        cin >> n ; 
        for(int i = 0; i < n; i++)
        {
            cin >> points[i].first >> points[i].second ;
            cin >> dir[i] ; 
        }
     
        ld cur, nxt, mid, nextMid, low = 0, high = 1e9 ;
        while((high - low) > PR)
        {
            mid = (low + high) / 2.0 ; 
            nextMid = mid + PR ;
            cur = maxArea(mid); nxt = maxArea(nextMid);
            
           // cout << high << " : " << low << " ::: " << cur << " - " << nxt << endl ; 
            if((cur - nxt) > PR) low = mid ;
            else high = mid  ; 
            // cout << high << " : " << low << endl ; 
        }
        cout << maxArea(high) ; 
    	return 0;
    }