#include <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define int long long
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
using namespace std;
typedef pair<int, int> P;
const int mod = 1000000007;
const int INF = 1e18;

double dx[4] = { 0, 1, 0, -1};
double dy[4] = {-1, 0, 1, 0};
string dir = "DRUL";

int n;
vector<double> x, y;
vector<char> d;

// double xx[4] = {}
// double yy[4] = {INF, 0, INF, 0}

double calc(double m){
    double MAXy = -INF, MAXx = -INF;
    double MINy = INF, MINx = INF;
    rep(i, 0, n){
        int idx = dir.find(d[i]);
        double nx = x[i] + m * dx[idx];
        double ny = y[i] + m * dy[idx];
        chmax(MAXx, nx);
        chmax(MAXy, ny);
        chmin(MINx, nx);
        chmin(MINy, ny);
    }
    return (MAXx - MINx) * (MAXy - MINy);
}

signed main(){
    cin >> n;
    x.resize(n); y.resize(n); d.resize(n);
    rep(i, 0, n){
        cin >> x[i] >> y[i] >> d[i];
    }
    double l = 0, r = 1e9;
    double MIN = INF;
    rep(i, 0, 200){
    // while(fabs(r - l) > 1e-10){
        double m1 = (l * 2 + r) / 3;
        double m2 = (l + r * 2) / 3;
        double t1 = calc(m1);
        double t2 = calc(m2);
        if(t1 < t2) r = m2;
        else l = m1;
        chmin(MIN, t1);
        chmin(MIN, t2);
    }
    printf("%.10f\n", MIN);
}
### 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 rep(i, a, n) for(int i = a; i < n; i++)
#define int long long
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
using namespace std;
typedef pair<int, int> P;
const int mod = 1000000007;
const int INF = 1e18;

double dx[4] = { 0, 1, 0, -1};
double dy[4] = {-1, 0, 1, 0};
string dir = "DRUL";

int n;
vector<double> x, y;
vector<char> d;

// double xx[4] = {}
// double yy[4] = {INF, 0, INF, 0}

double calc(double m){
    double MAXy = -INF, MAXx = -INF;
    double MINy = INF, MINx = INF;
    rep(i, 0, n){
        int idx = dir.find(d[i]);
        double nx = x[i] + m * dx[idx];
        double ny = y[i] + m * dy[idx];
        chmax(MAXx, nx);
        chmax(MAXy, ny);
        chmin(MINx, nx);
        chmin(MINy, ny);
    }
    return (MAXx - MINx) * (MAXy - MINy);
}

signed main(){
    cin >> n;
    x.resize(n); y.resize(n); d.resize(n);
    rep(i, 0, n){
        cin >> x[i] >> y[i] >> d[i];
    }
    double l = 0, r = 1e9;
    double MIN = INF;
    rep(i, 0, 200){
    // while(fabs(r - l) > 1e-10){
        double m1 = (l * 2 + r) / 3;
        double m2 = (l + r * 2) / 3;
        double t1 = calc(m1);
        double t2 = calc(m2);
        if(t1 < t2) r = m2;
        else l = m1;
        chmin(MIN, t1);
        chmin(MIN, t2);
    }
    printf("%.10f\n", MIN);
}
```