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 <vector>
#include <algorithm>
#include <string>
#include <iomanip>
#include <cmath>
using namespace std;
typedef long long ll;
string direction = "RLUD";
int dx[4] = {1,-1,0,0},dy[4] = {0,0,1,-1};

int N;
int main(){
    cin >> N;
    vector<ll> X[4],Y[4];
    ll x,y; char c;
    for(int i=0;i<N;i++){
        cin >> x >> y >> c;
        x *= 2; y *= 2;
        int d = direction.find(c);
        X[d].push_back(x); Y[d].push_back(y);
    }
    vector<ll> xs,ys;
    for(int k=0;k<4;k++){//各方向のx,yの最大値、最小値を入れる
        if(X[k].size()){
            xs.push_back(*max_element(X[k].begin(),X[k].end()));
            xs.push_back(*min_element(X[k].begin(),X[k].end()));
            ys.push_back(*max_element(Y[k].begin(),Y[k].end()));
            ys.push_back(*min_element(Y[k].begin(),Y[k].end()));
        }
    }
    vector<ll> t;
    for(auto x1:xs) for(auto x2:xs){//最適解を達成する時刻を列挙
        ll d = abs(x1-x2);
        t.push_back(d); t.push_back(d/2);
    }
    for(auto y1:ys) for(auto y2:ys){
        ll d = abs(y1-y2);
        t.push_back(d); t.push_back(d/2);
    }
    sort(t.begin(),t.end());
    t.erase(unique(t.begin(),t.end()),t.end());
    ll inf = 1e9,ans = 1e18;
    for(auto d:t){//時刻の候補をすべて試す
        ll xmax = -inf,xmin = inf,ymax = -inf,ymin = inf;
        for(int k=0;k<4;k++) for(int i=0;i<X[k].size();i++){
            ll x = X[k][i]+dx[k]*d,y = Y[k][i]+dy[k]*d;
            xmax = max(xmax,x);
            xmin = min(xmin,x);
            ymax = max(ymax,y);
            ymin = min(ymin,y);
        }
        ans = min(ans,(xmax-xmin)*(ymax-ymin));
    }
    cout << fixed;
    cout << setprecision(10) << 1.0*ans/4 << endl;
}