#include <bits/stdc++.h>

using namespace std;

#ifdef DGC
#include "debug.h"
#else
#define debug(...) 9715
#endif
typedef long long ll;
typedef long double ld;
typedef complex<ll> point;
#define F first
#define S second



int main()
{
	#ifdef DGC
		freopen("a.in", "r", stdin);
		//freopen("a.out", "w", stdout);
	#endif

	ios_base::sync_with_stdio(0), cin.tie(0);

	int n;
	cin >> n;
	map<char, vector<pair<int, int>>> mp;
	char z;
	for (int i = 0, x, y; i < n; ++i)
	{
		cin >> x >> y >> z;
		mp[z].push_back({ x, y });
	}

	int mnx = 1e9, mxx = -1e9, mny = 1e9, mxy = -1e9;
	for (auto i : mp['L']) mny = min(mny, i.S), mxy = max(mxy, i.S);
	for (auto i : mp['R']) mny = min(mny, i.S), mxy = max(mxy, i.S);
	for (auto i : mp['U']) mnx = min(mnx, i.F), mxx = max(mxx, i.F);
	for (auto i : mp['D']) mnx = min(mnx, i.F), mxx = max(mxx, i.F);

	auto f1 = [&](ld md)
	{
		ld fmn = mnx, fmx = mxx;
		for (auto i : mp['L'])
		{
			fmn = min(fmn, i.F - md);
			fmx = max(fmx, i.F - md);
		}
		for (auto i : mp['R'])
		{
			fmn = min(fmn, i.F + md);
			fmx = max(fmx, i.F + md);
		}
		fmn = min(fmn, fmx);
		return fmx-fmn;
	};

	auto f2 = [&](ld md)
	{
		ld fmn = mny, fmx = mxy;
		for (auto i : mp['D'])
		{
			fmn = min(fmn, i.S - md);
			fmx = max(fmx, i.S - md);
		}
		for (auto i : mp['U'])
		{
			fmn = min(fmn, i.S + md);
			fmx = max(fmx, i.S + md);
		}
		fmn = min(fmn, fmx);
		return fmx-fmn;
	};

	ld l = 0, r = 1e9+5;
	for (int it = 200; it--; )
	{
		ld md1 = l + (r-l) / 3.0;
		ld md2 = r - (r-l) / 3.0;
		if (f1(md1) * f2(md1) < f1(md2) * f2(md2))
			r = md2;
		else
			l = md1;
	}

	cout << setprecision(20) << fixed << f1(l) * f2(l) << "\n";

	return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;

#ifdef DGC
#include "debug.h"
#else
#define debug(...) 9715
#endif
typedef long long ll;
typedef long double ld;
typedef complex<ll> point;
#define F first
#define S second



int main()
{
	#ifdef DGC
		freopen("a.in", "r", stdin);
		//freopen("a.out", "w", stdout);
	#endif

	ios_base::sync_with_stdio(0), cin.tie(0);

	int n;
	cin >> n;
	map<char, vector<pair<int, int>>> mp;
	char z;
	for (int i = 0, x, y; i < n; ++i)
	{
		cin >> x >> y >> z;
		mp[z].push_back({ x, y });
	}

	int mnx = 1e9, mxx = -1e9, mny = 1e9, mxy = -1e9;
	for (auto i : mp['L']) mny = min(mny, i.S), mxy = max(mxy, i.S);
	for (auto i : mp['R']) mny = min(mny, i.S), mxy = max(mxy, i.S);
	for (auto i : mp['U']) mnx = min(mnx, i.F), mxx = max(mxx, i.F);
	for (auto i : mp['D']) mnx = min(mnx, i.F), mxx = max(mxx, i.F);

	auto f1 = [&](ld md)
	{
		ld fmn = mnx, fmx = mxx;
		for (auto i : mp['L'])
		{
			fmn = min(fmn, i.F - md);
			fmx = max(fmx, i.F - md);
		}
		for (auto i : mp['R'])
		{
			fmn = min(fmn, i.F + md);
			fmx = max(fmx, i.F + md);
		}
		fmn = min(fmn, fmx);
		return fmx-fmn;
	};

	auto f2 = [&](ld md)
	{
		ld fmn = mny, fmx = mxy;
		for (auto i : mp['D'])
		{
			fmn = min(fmn, i.S - md);
			fmx = max(fmx, i.S - md);
		}
		for (auto i : mp['U'])
		{
			fmn = min(fmn, i.S + md);
			fmx = max(fmx, i.S + md);
		}
		fmn = min(fmn, fmx);
		return fmx-fmn;
	};

	ld l = 0, r = 1e9+5;
	for (int it = 200; it--; )
	{
		ld md1 = l + (r-l) / 3.0;
		ld md2 = r - (r-l) / 3.0;
		if (f1(md1) * f2(md1) < f1(md2) * f2(md2))
			r = md2;
		else
			l = md1;
	}

	cout << setprecision(20) << fixed << f1(l) * f2(l) << "\n";

	return 0;
}

```