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:
/* 
 * Author: aps
 * Time: 19-06-2019 00:54:25
**/
#include<bits/stdc++.h>

#define fi first
#define se second
#define pb push_back
#define all(x) x.begin(), x.end()

using namespace std;

typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;
typedef vector<int> vi;
typedef pair<int,int> pii;

const ll inf = 1e18;
const ll mod = 1e9+7;

inline ll add(ll x, ll y) { x += y; if (x >= mod) x -= mod; return x;}
inline ll sub(ll x, ll y) { x -= y; if (x < 0) x += mod; return x;}
inline ll mul(ll x, ll y) { return ((x % mod) * (y % mod)) % mod;}
inline ll power(ll a, ll b) {
    ll x = 1;
    while (b) {
        if (b & 1) x = mul(x, a);
        a = mul(a, a);
        b >>= 1;
    }
    return x;
}
inline ll inv(ll a) { return power(a, mod - 2);}

const int N = 1e5+5;

ld x[N],y[N];
char dir[N];
int n;

ld f(ld t){
    ld xmin = INT_MAX, xmax = INT_MIN, ymin = INT_MAX, ymax = INT_MIN;
    for(int i=0;i<n;i++){
        ld cx = x[i], cy = y[i];
        if(dir[i] == 'U') cy += t;
        if(dir[i] == 'D') cy -= t;
        if(dir[i] == 'R') cx += t;
        if(dir[i] == 'L') cx -= t;
        
        xmin = min(xmin,cx);
        xmax = max(xmax,cx);

        ymin = min(ymin,cy);
        ymax = max(ymax,cy);
    }
   return (ymax-ymin)*(xmax-xmin); 
}

int main(){
    ios_base::sync_with_stdio(false); cin.tie(0);
    
    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(f(m1) < f(m2)){
            r = m2;
        } else {
            l = m1;
        }
    }

    cout << fixed << setprecision(9) <<  f(l) << endl;


    return 0;
}