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:
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("fma,avx,avx2")

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define ld long double
#define sz(a) (int) (a).size()
#define all(a) (a).begin(), (a).end()
#define x first
#define y second

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

const ld inf = 6e8;
const ld eps = 1e-10;

int n;
vector<ld> x, y;
vector<char> d;

ld get_x(int i, ld t) {
    if (d[i] == 'L') return x[i] - t;
    if (d[i] == 'R') return x[i] + t;
    return x[i];
}

ld get_y(int i, ld t) {
    if (d[i] == 'D') return y[i] - t;
    if (d[i] == 'U') return y[i] + t;
    return y[i];
}

ld solve(ld t) {
    ld mx_x = -inf * inf, mn_x = inf * inf;
    ld mx_y = -inf * inf, mn_y = inf * inf;
    for (int i = 0; i < n; i++) {
        ld xx = get_x(i, t);
        ld yy = get_y(i, t);
        mx_x = max(mx_x, xx);
        mn_x = min(mn_x, xx);
        mx_y = max(mx_y, yy);
        mn_y = min(mn_y, yy);
    }
    return (mx_x - mn_x) * (mx_y - mn_y);
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    cin >> n;
    x.resize(n);
    y.resize(n);
    d.resize(n);
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> y[i] >> d[i];
    }
    ld l = 0.0, r = inf, m2, m1;
    ld three = 3.0;
    for (int i = 0; i < 600; i++) {
        m1 = l + (r - l) / three;
        m2 = r - (r - l) / three;
        if (solve(m1) + eps > solve(m2)) {
            l = m1;
        } else {
            r = m2;
        }
    }
    cout << fixed << setprecision(12) << solve(l) << "\n";
    return 0;
}