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<iomanip>
#include<vector>
#include<string>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<climits>
#include<fstream>

#define MAX_N (100000+10)
#define MOD (1000000007)

using namespace std;

typedef long long int Int;

constexpr Int TEN(int n) { return n == 0 ? 1 : 10 * TEN(n-1); }

Int N;
const double inf = 1e18;
double X[MAX_N],Y[MAX_N],Dx[MAX_N],Dy[MAX_N];

double func(double a) {
    double minx = inf,maxx = -inf, miny = inf,maxy = -inf;

    for(int i = 0;i < N;i++) {
        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 solve(double l, double r) {
    double ans = inf;
    int num = 300;
    while(num--) {
        double left = l + (r - l) / 3;
        double right = r - (r - l) / 3;

        double f1 = func(left);
        double f2 = func(right);

        ans = min(ans,min(f1,f2));

        if(f1 < f2)
            r = right;
        else
            l = left;
    }

    return ans;
}

int main(void) {
    cin>>N;
    char s;

    for(int i = 0;i < N;i++) {
        cin>>X[i]>>Y[i]>>s;
        if(s == 'R')
            Dx[i] = 1;
        else if(s == 'L')
            Dx[i] = -1;
        else if(s == 'U')
            Dy[i] = 1;
        else
            Dy[i] = -1;
    }

    printf("%f\n",solve(0, 1e9));

    return 0;
}