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;

#define watch(x) cerr << (#x) << ": " << (x) << endl
#define int long long

struct point {
    double x, y;
    char d;
    point(int _x, int _y, char _d) {
        x = _x; y = _y; d = _d;
    }
};

vector<point> points;

double score(double time) {
    double mxx = -1e9, mnx = 1e9, mxy = -1e9, mny = 1e9;

    for(point p : points) {
        if(p.d == 'U') {
            mxx = max(mxx, p.x);
            mnx = min(mnx, p.x);
            mxy = max(mxy, p.y+time);
            mny = min(mny, p.y+time);
        } else if(p.d == 'D') {
            mxx = max(mxx, p.x);
            mnx = min(mnx, p.x);
            mxy = max(mxy, p.y-time);
            mny = min(mny, p.y-time);
        } else if(p.d == 'R') {
            mxx = max(mxx, p.x+time);
            mnx = min(mnx, p.x+time);
            mxy = max(mxy, p.y);
            mny = min(mny, p.y);
        } else {
            mxx = max(mxx, p.x-time);
            mnx = min(mnx, p.x-time);
            mxy = max(mxy, p.y);
            mny = min(mny, p.y);
        }
    }

    return (mxx-mnx)*(mxy-mny);
}

int32_t main() {

    int n; cin >> n;
    
    for(int i = 0; i < n; ++i) {
        double x, y; cin >> x >> y;
        char c; cin >> c;

        points.emplace_back(x, y, c);
    }

    double l = 0, r = 1e9;

    for(int i = 0; i < 500; ++i) {
        double lmid = (l+l+r)/3.0, rmid = (l+r+r)/3.0;

        if(score(lmid) < score(rmid)) {
            r = rmid;
        } else {
            l = lmid;
        }
    }

    cout << setprecision(10) << score(l) << endl;
    
    return 0;
}