#include <bits/stdc++.h>
#define For(i, a, b) for(int (i)=(a); (i)<(b); ++(i))
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef pair<int, int> pii;

int n;
double x[100010], y[100010];
char d[100010];

double f(double t){
    vector<double> tx(n), ty(n);
    rep(i, n){
        tx[i] = x[i];
        ty[i] = y[i];
        if(d[i] == 'L') tx[i] -= t;
        if(d[i] == 'R') tx[i] += t;
        if(d[i] == 'U') ty[i] += t;
        if(d[i] == 'D') ty[i] -= t;
    }
    sort(tx.begin(), tx.end());
    sort(ty.begin(), ty.end());
    return (tx[n-1] - tx[0]) * (ty[n-1] - ty[0]);
}

int main(){
    scanf("%d", &n);
    rep(i, n) cin >> x[i] >> y[i] >> d[i];
    double high = 1e9;
    double low = 0;
    double m2 = high;
    double m1 = low;
    rep(i, 100){
        m1 = (high + low*2) / 3;
        m2 = (high*2 + low) / 3;
        if(f(m1) > f(m2)) low = m1;
        else high = m2;
    }
    printf("%.10f", min({f(low), f(m1), f(m2), f(high)}));
}
### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define For(i, a, b) for(int (i)=(a); (i)<(b); ++(i))
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef pair<int, int> pii;

int n;
double x[100010], y[100010];
char d[100010];

double f(double t){
    vector<double> tx(n), ty(n);
    rep(i, n){
        tx[i] = x[i];
        ty[i] = y[i];
        if(d[i] == 'L') tx[i] -= t;
        if(d[i] == 'R') tx[i] += t;
        if(d[i] == 'U') ty[i] += t;
        if(d[i] == 'D') ty[i] -= t;
    }
    sort(tx.begin(), tx.end());
    sort(ty.begin(), ty.end());
    return (tx[n-1] - tx[0]) * (ty[n-1] - ty[0]);
}

int main(){
    scanf("%d", &n);
    rep(i, n) cin >> x[i] >> y[i] >> d[i];
    double high = 1e9;
    double low = 0;
    double m2 = high;
    double m1 = low;
    rep(i, 100){
        m1 = (high + low*2) / 3;
        m2 = (high*2 + low) / 3;
        if(f(m1) > f(m2)) low = m1;
        else high = m2;
    }
    printf("%.10f", min({f(low), f(m1), f(m2), f(high)}));
}
```