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: Accepted)

#include <iostream>
#include <algorithm>
#include <utility>
#include <vector>
#include <string>
#include <iomanip>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define REP(i, n) for (int i = 0; i < (n); i++)
#define RREP(i, n) for (int i = (n) - 1; i >= 0; i--)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define RFOR(i, a, b) for (int i = (a); i > (b); i--)
#define ALL(a) (a).begin(), (a).end()
#define INF 1e10

struct D {
    vector<double> l, r;
    D() : l(3, INF), r(3, -INF) {}

    void add(double x, int v) {
        v++;
        l[v] = min(l[v], x);
        r[v] = max(r[v], x);
    }

    double get(double t) {
        double nl = INF, nr = -INF;
        REP(i, 3) {
            nl = min(nl, l[i] + i * t);
            nr = max(nr, r[i] + i * t);
        }
        return nr - nl;
    }

    vector<double> events() {
        vector<double> ret;
        FOR(i, 1, 3) REP(j, i) {
            double t = (l[j] - l[i]) / (i - j);
            if (t > 0) ret.push_back(t);
            t = (r[j] - r[i]) / (i - j);
            if (t > 0) ret.push_back(t);
        }
        return ret;
    }
};

int main() {
    int N;
    cin >> N;
    D xd, yd;
    REP(i, N) {
        int x, y;
        char d;
        cin >> x >> y >> d;
        switch (d) {
            case 'L': xd.add(x, -1), yd.add(y, 0); break;
            case 'R': xd.add(x, 1), yd.add(y, 0); break;
            case 'U': xd.add(x, 0), yd.add(y, 1); break;
            case 'D': xd.add(x, 0), yd.add(y, -1); break;
        }
    }

    vector<double> ts;
    ts.push_back(0);
    ts.push_back(INF);

    vector<double> nts = xd.events();
    ts.insert(ts.end(), ALL(nts));
    nts = yd.events();
    ts.insert(ts.end(), ALL(nts));
    sort(ALL(ts));

    double ans = 1e18;
    for (auto t : ts) {
        ans = min(ans, xd.get(t) * yd.get(t));
    }
    cout << setprecision(12) << ans << endl;
}