#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5, mod = 1e9 + 7;
const double inf = 1e18;
struct e{
    int x, y;
    double nx, ny;
    char p;
}a[maxn];
int n;
double F(double mid)
{
    double maxx = -inf, minx = inf, maxy = -inf, miny = inf;
    for (int i = 0; i < n; ++i){
        a[i].nx = a[i].x, a[i].ny = a[i].y;
        if(a[i].p == 'L') a[i].nx = (double)a[i].x - mid;
        else if(a[i].p == 'R') a[i].nx = (double)a[i].x + mid;
        else if(a[i].p == 'U') a[i].ny = (double)a[i].y + mid;
        else a[i].ny = (double)a[i].y - mid;
        minx = min(minx, a[i].nx);
        maxx = max(maxx, a[i].nx);
        miny = min(miny, a[i].ny);
        maxy = max(maxy, a[i].ny);
    }
    return (maxx - minx) * (maxy - miny);
}
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; ++i){
        cin >> a[i].x >> a[i].y >> a[i].p;
    }
    double l = 0, r = 1e8;
    for (int i = 0; i < 100; ++i){
        double len = (r - l) / 3;
        double lmid = l + len, rmid = l + 2 * len;
        if (F(lmid) < F(rmid)) r = rmid;
        else l = lmid;
    }
    cout << fixed << F((l + r) / 2) << endl;
    return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5, mod = 1e9 + 7;
const double inf = 1e18;
struct e{
    int x, y;
    double nx, ny;
    char p;
}a[maxn];
int n;
double F(double mid)
{
    double maxx = -inf, minx = inf, maxy = -inf, miny = inf;
    for (int i = 0; i < n; ++i){
        a[i].nx = a[i].x, a[i].ny = a[i].y;
        if(a[i].p == 'L') a[i].nx = (double)a[i].x - mid;
        else if(a[i].p == 'R') a[i].nx = (double)a[i].x + mid;
        else if(a[i].p == 'U') a[i].ny = (double)a[i].y + mid;
        else a[i].ny = (double)a[i].y - mid;
        minx = min(minx, a[i].nx);
        maxx = max(maxx, a[i].nx);
        miny = min(miny, a[i].ny);
        maxy = max(maxy, a[i].ny);
    }
    return (maxx - minx) * (maxy - miny);
}
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; ++i){
        cin >> a[i].x >> a[i].y >> a[i].p;
    }
    double l = 0, r = 1e8;
    for (int i = 0; i < 100; ++i){
        double len = (r - l) / 3;
        double lmid = l + len, rmid = l + 2 * len;
        if (F(lmid) < F(rmid)) r = rmid;
        else l = lmid;
    }
    cout << fixed << F((l + r) / 2) << endl;
    return 0;
}

```