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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define mins(x,y) x = min(x,y)
#define maxs(x,y) x = max(x,y)
using namespace std;

const double inf = 1e10;

struct D {
    vector<double> l, r;
    D():l(3,inf),r(3,-inf) {}
    void add(double x, int v) {
        ++v;
        mins(l[v],x);
        maxs(r[v],x);
    }
    double get(double t) {//幅
        double nl = inf, nr = -inf;
        rep(i,3) {
            mins(nl, l[i]+i*t);
            maxs(nr, r[i]+i*t);
        }
        return nr-nl;
    }
    vector<double> events() {
        vector<double> ts;
        rep(i,3)rep(j,i) { // j < i
            {
            double t = (l[j]-l[i])/(i-j);
            if (t > 0) ts.push_back(t);
            }
            {
            double t = (r[j]-r[i])/(i-j);
            if (t > 0) ts.push_back(t);
            }
        }
        return ts;
    }
};

int main() {
    int n;
    cin >> n;
    D xd, yd;
    rep(i,n) {
        int x, y;
        char d;
        cin >> x >> y >> d;
        if (d == 'L') xd.add(x,-1), yd.add(y,0);
        if (d == 'R') xd.add(x,1), yd.add(y,0);
        if (d == 'U') xd.add(x,0), yd.add(y,1);
        if (d == 'D') xd.add(x,0), yd.add(y,-1);
    }
    vector<double> ts;
    ts.push_back(0);
    //ts.push_back(inf);
    {
    auto nts = xd.events();
    ts.insert(ts.end(),nts.begin(),nts.end());
    }
    {
    auto nts = yd.events();
    ts.insert(ts.end(),nts.begin(),nts.end());
    }
    sort(ts.begin(),ts.end());
    double ans = 1e18;
    for (double t : ts) {
        double now = xd.get(t)*yd.get(t);
        mins(ans,now);
    }
    printf("%.10f\n", ans);
    return 0;
}