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 <iostream>
#include <algorithm>
#include <iomanip>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using u32 = uint32_t;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;

int main() {
    int n;
    cin >> n;
    vector<int> x(n), y(n);
    vector<char> d(n);
    for (int i = 0; i < n; ++i) {
        cin >> x[i] >> y[i] >> d[i];
    }
    double l = 0, r = INF<int>;
    double xmax = INF<ll>, xmin = -INF<ll>, ymax = INF<ll>, ymin = -INF<ll>;
    double xmax2 = INF<ll>, xmin2 = -INF<ll>, ymax2 = INF<ll>, ymin2 = -INF<ll>;
    for (int i = 0; i < 300; ++i) {
        double lb = (l*2+r)/3, ub = (l+2*r)/3;
        xmax = -INF<ll>, xmin = INF<ll>, ymax = -INF<ll>, ymin = INF<ll>;
        xmax2 = -INF<ll>, xmin2 = INF<ll>, ymax2 = -INF<ll>, ymin2 = INF<ll>;
        for (int j = 0; j < n; ++j) {
            double X1 = x[j], Y1 = y[j], X2 = x[j], Y2 = y[j];
            if(d[j] == 'R')  X1 += lb,  X2 += ub;
            if(d[j] == 'L')  X1 -= lb,  X2 -= ub;
            if(d[j] == 'U')  Y1 += lb,  Y2 += ub;
            if(d[j] == 'D')  Y1 -= lb,  Y2 -= ub;
            xmax = max(xmax, X1); xmin = min(xmin, X1);
            ymax = max(ymax, Y1); ymin = min(ymin, Y1);
            xmax2 = max(xmax2, X2); xmin2 = min(xmin2, X2);
            ymax2 = max(ymax2, Y2); ymin2 = min(ymin2, Y2);
        }
        if((xmax-xmin)*(ymax-ymin) < (xmax2-xmin2)*(ymax2-ymin2)) r = ub;
        else l = lb;
    }
    cout << setprecision(15) << (xmax-xmin)*(ymax-ymin) << "\n";
    return 0;
}