#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
#define double ld
typedef complex<ld> point;
#define F first
#define S second
#define ii pair<int,int>
#define iii pair<int,ii>



int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);

    int n;
    cin>>n;
    vector<ii> v;
    vector<string> w;
    for(int i=0; i<n; i++)
    {
        int x, y;
        string s;
        cin>>x>>y>>s;
        v.push_back(ii(x, y));
        w.push_back(s);
    }

    auto val = [&](ld x)
    {
        ld lx = 1e9, rx = -1e9, ly = 1e9, ry = -1e9;
        for(int i=0; i<n; i++)
        {
            ld x1 = v[i].F, y1 = v[i].S;
            if(w[i] == "U")
                y1 += x;
            if(w[i] == "D")
                y1 -= x;
            if(w[i] == "R")
                x1 += x;
            if(w[i] == "L")
                x1 -= x;
            lx = min(lx, x1);
            rx = max(rx, x1);
            ly = min(ly, y1);
            ry = max(ry, y1);
        }
        return (rx - lx) * (ry - ly);
    };

    ld l = 0, r = 2e8, ans = 1e18;
    for(int i=0; i<200; i++)
    {
        ld mid1 = l + (r - l) / 3;
        ld val1 = val(mid1);

        ld mid2 = l + 2.0 * (r - l) / 3;
        ld val2 = val(mid2);

        if(val1 < val2)
            r = mid2;
        else
            l = mid1;

        ans = min(ans, min(val1, val2));
    }

    cout<<setprecision(10)<<fixed;
    cout<<ans<<'\n';



    return 0;
}

### Prompt

Create a solution in CPP for 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>
using namespace std;

typedef long long ll;
typedef long double ld;
#define double ld
typedef complex<ld> point;
#define F first
#define S second
#define ii pair<int,int>
#define iii pair<int,ii>



int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);

    int n;
    cin>>n;
    vector<ii> v;
    vector<string> w;
    for(int i=0; i<n; i++)
    {
        int x, y;
        string s;
        cin>>x>>y>>s;
        v.push_back(ii(x, y));
        w.push_back(s);
    }

    auto val = [&](ld x)
    {
        ld lx = 1e9, rx = -1e9, ly = 1e9, ry = -1e9;
        for(int i=0; i<n; i++)
        {
            ld x1 = v[i].F, y1 = v[i].S;
            if(w[i] == "U")
                y1 += x;
            if(w[i] == "D")
                y1 -= x;
            if(w[i] == "R")
                x1 += x;
            if(w[i] == "L")
                x1 -= x;
            lx = min(lx, x1);
            rx = max(rx, x1);
            ly = min(ly, y1);
            ry = max(ry, y1);
        }
        return (rx - lx) * (ry - ly);
    };

    ld l = 0, r = 2e8, ans = 1e18;
    for(int i=0; i<200; i++)
    {
        ld mid1 = l + (r - l) / 3;
        ld val1 = val(mid1);

        ld mid2 = l + 2.0 * (r - l) / 3;
        ld val2 = val(mid2);

        if(val1 < val2)
            r = mid2;
        else
            l = mid1;

        ans = min(ans, min(val1, val2));
    }

    cout<<setprecision(10)<<fixed;
    cout<<ans<<'\n';



    return 0;
}

```