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 string LRDU = "LRDU";
int xsig[] = {0, 2, 1, 1};
int ysig[] = {1, 1, 0, 2};

void chmax(int64_t& a, int64_t b){
    a = max(a, b);
}
void chmin(int64_t& a, int64_t b){
    a = min(a, b);
}

int main(){
    int N;
    cin >> N;
    vector<int64_t> Xmin(3, 1e9), Xmax(3, -1e9), Ymin(3, 1e9), Ymax(3, -1e9);
    for(int i=0; i<N; i++){
        int x, y;
        char d;
        cin >> x >> y >> d;
        x *= 2;
        y *= 2;
        int c = LRDU.find(d);
        chmin(Xmin[xsig[c]], x);
        chmax(Xmax[xsig[c]], x);
        chmin(Ymin[ysig[c]], y);
        chmax(Ymax[ysig[c]], y);
    }
    vector<int64_t> xs, ys;
    for(int64_t a : Xmin) if(abs(a) < 1e9) xs.push_back(a);
    for(int64_t a : Xmax) if(abs(a) < 1e9) xs.push_back(a);
    for(int64_t a : Ymin) if(abs(a) < 1e9) ys.push_back(a);
    for(int64_t a : Ymax) if(abs(a) < 1e9) ys.push_back(a);

    vector<int64_t> ts;
    for(int64_t a : xs) for(int64_t b : xs){
        ts.push_back(abs(a-b));
        ts.push_back(abs(a-b)/2);
    }
    for(int64_t a : ys) for(int64_t b : ys){
        ts.push_back(abs(a-b));
        ts.push_back(abs(a-b)/2);
    }

    int64_t ans = 1e18;
    for(int64_t t : ts){
        int64_t xmin = min({Xmin[0]-t, Xmin[1], Xmin[2]+t});
        int64_t xmax = max({Xmax[0]-t, Xmax[1], Xmax[2]+t});
        int64_t ymin = min({Ymin[0]-t, Ymin[1], Ymin[2]+t});
        int64_t ymax = max({Ymax[0]-t, Ymax[1], Ymax[2]+t});
        chmin(ans, (xmax-xmin) * (ymax-ymin));
    }
    cout << fixed << setprecision(10) << double(ans)/4 << endl;
}