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:
#include <bits/stdc++.h>
#include <math.h>
using namespace std;
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
typedef long double ld;
typedef long long ll;

struct Point {
    ll x, y, x1 = 0, y1 = 0;
    
    Point() {}
    
    Point(ll _x, ll _y, char c) {
        x = _x;
        y = _y;
        if (c == 'U') {
            y1 = 1;
        } else if (c == 'D') {
            y1 = -1;
        } else if (c == 'R') {
            x1 = 1;
        } else {
            x1 = -1;
        }
    }
};

vector<Point> vc;

ld dist(ld t) {
    ld max_x = -1e9, max_y = -1e9, min_x = 1e9, min_y = 1e9;
    for (auto p : vc) {
        ld xx = p.x + p.x1 * t;
        ld yy = p.y + p.y1 * t;
        max_x = max(max_x, xx);
        max_y = max(max_y, yy);
        min_x = min(min_x, xx);
        min_y = min(min_y, yy);
    }
    return (max_x - min_x) * (max_y - min_y);
}

int main() {
#ifdef LOCAL
    freopen("in", "r", stdin);
    freopen("out", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.precision(20);
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        ll x, y;
        char c;
        cin >> x >> y >> c;
        vc.push_back(Point(x, y, c));
    }
    ld l = 0, r = 1e11;
    for (int i = 0; i < 700; ++i) {
        ld m1 = (r - l) / 3 + l;
        ld m2 = r - (r - l) / 3;
        ld a1 = dist(m1);
        ld a2 = dist(m2);
        if (a1 > a2) {
            l = m1;
        } else {
            r = m2;
        }
    }
    cout << fixed << min(dist(l), dist(r));
    return 0;
}