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 INF 1<<30
#define endl '\n'
#define maxn 100005
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
typedef long double ld;
const double PI = acos(-1.0);
#define dbg(x) cerr << #x << " = " << x << endl;
#define dbg2(x, y) cerr << #x << " = " << x << ", " << #y << " = " << y << endl;
#define dbg3(x, y, z) cerr << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl;

const ll mod = 1e9 + 7;
ld X[maxn], Y[maxn];
char dir[maxn];
int n;

ld solve(ld t)
{
    ld x_mn = INT_MAX, x_mx  = INT_MIN, y_mn = INT_MAX, y_mx = INT_MIN;
    for(int i = 0; i < n; i++){
        ld cx = X[i], cy = Y[i];
        if(dir[i] == 'U') cy +=  t;
        else if(dir[i] == 'D') cy -= t;
        else if(dir[i] == 'R') cx += t;
        else if(dir[i] == 'L') cx -= t;

        x_mn = min(x_mn, cx);
        x_mx = max(x_mx, cx);

        y_mn = min(y_mn, cy);
        y_mx = max(y_mx, cy); 
    }
    return (y_mx - y_mn) * (x_mx - x_mn);
}

int main()
{
    FASTIO
    /*
    //double start_time = clock();
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    freopen("error.txt", "w", stderr);
#endif
//*/
    cin >> n;
    for(int i = 0; i < n; i++){
        cin >> X[i] >> Y[i] >> dir[i];
    }

    ld l = 0.0, r = 1e9;
    for(int i = 0; i < 300; i++){
        ld m1 = l + (r -l)/3.0;
        ld m2 = r - (r - l)/3.0;

        if(solve(m1) < solve(m2)) r  = m2;
        else l = m1;
    }
    cout << fixed << setprecision(9) << solve(l)<<endl;
    //double end_time = clock();
    //printf( "Time = %lf ms\n", ( (end_time - start_time) / CLOCKS_PER_SEC)*1000);

    return 0;
}