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>
using namespace std;

const int N = 1e5 + 5;
const double inf = 1e18;
double x[N], y[N];
double dx[N], dy[N];
int n;

double f(double a) {
    double minx = inf, maxx = -inf, miny = inf, maxy = -inf;
    for (int i = 0; i < n; i++) { //n个点
        double nx = x[i] + a * dx[i];
        double ny = y[i] + a * dy[i];
        minx = min(minx, nx);
        maxx = max(maxx, nx);
        miny = min(miny, ny);
        maxy = max(maxy, ny);
    }
    return (maxx - minx) * (maxy - miny);
}

double ternary_search(double l, double r) {
   double ans = inf;
  int times = 300;
    while (times--) {
        double m1 = l + (r - l) / 3;
         double m2 = r - (r - l) / 3;
        //double m1 = (l*2+r) / 3;
        //double m2 = (r*2+l) / 3;
        double f1 = f(m1);
        double f2 = f(m2);
        ans = min(ans, f1);
         ans = min(ans, f2);
        if (f1 < f2) r = m2;
        else l = m1;
    }
    return ans;
}

int main() {
   // while (cin >> n) {
      cin>>n;
        //memset(dx, 0, sizeof(dx));
        //memset(dy, 0, sizeof(dy));

        string s;
        for (int i = 0; i < n; i++) {
            cin >> x[i] >> y[i] >> s;
            if (s == "R") dx[i] = 1;
            if (s == "L") dx[i] = -1;
            if (s == "U") dy[i] = 1;
            if (s == "D") dy[i] = -1;
        }


        //cout << ternary_search(0, 1e9) << endl;
      printf("%lf\n",ternary_search(0, 1e9));
   // }

    return 0;
}