#pragma once

#include <bits/stdc++.h>

#define M_PI       3.14159265358979323846   // pi

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> t3;

#define rep(a,n) for(int a = 0;a < n;a++)
#define repi(a,b,n) for(int a = b;a < n;a++)

const ull mod = 1000000007;

template<typename T>
struct TriSearch {
	T solve(T left, T right, function<T(T)> pred, function<bool(T, T)> condition) {
		while (condition(left, right))
		{
			T lmid = (left * 2 + right) / 3;
			T rmid = (left + right * 2) / 3;
			T L = pred(lmid);
			T R = pred(rmid);
			if (L > R) left = lmid;
			else right = rmid;
		}
		return right;

	}
};


int main(void)
{
	ll n;
	cin >> n;
	vector<vector<ll>> vs(n, vector<ll>(4,0));
	rep(i, n) {
		char c;
		cin >> vs[i][0] >> vs[i][1] >> c;
		if (c == 'U') 
		{
			vs[i][2] = 0;
			vs[i][3] = 1;
		}
		else if (c == 'D')
		{
			vs[i][2] = 0;
			vs[i][3] = -1;
		}
		else if (c == 'L')
		{
			vs[i][2] = -1;
			vs[i][3] = 0;
		}
		else
		{
			vs[i][2] = 1;
			vs[i][3] = 0;
		}
	}
	TriSearch<double> t;
	double area = 2e18;
	int count = 0;
	double r = t.solve(0, 1e9, [&](double v) {
		double xmin = 1e18, xmax = -1e18, ymin = 1e18, ymax = -1e18;
		for (auto e : vs) {
			double x = e[0] + e[2] * v;
			double y = e[1] + e[3] * v;
			xmin = min(xmin, x);
			xmax = max(xmax, x);
			ymin = min(ymin, y);
			ymax = max(ymax, y);
		}
		auto a = (xmax - xmin) * (ymax - ymin);
		area = min(a, area);
		return a;
		}, [&](double l, double r) {
			count++;
			return count < 150;
		});
	cout << std::fixed << std::setprecision(15) << area << endl;
	return 0;
}

### Prompt

Please create a solution in cpp 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
#pragma once

#include <bits/stdc++.h>

#define M_PI       3.14159265358979323846   // pi

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> t3;

#define rep(a,n) for(int a = 0;a < n;a++)
#define repi(a,b,n) for(int a = b;a < n;a++)

const ull mod = 1000000007;

template<typename T>
struct TriSearch {
	T solve(T left, T right, function<T(T)> pred, function<bool(T, T)> condition) {
		while (condition(left, right))
		{
			T lmid = (left * 2 + right) / 3;
			T rmid = (left + right * 2) / 3;
			T L = pred(lmid);
			T R = pred(rmid);
			if (L > R) left = lmid;
			else right = rmid;
		}
		return right;

	}
};


int main(void)
{
	ll n;
	cin >> n;
	vector<vector<ll>> vs(n, vector<ll>(4,0));
	rep(i, n) {
		char c;
		cin >> vs[i][0] >> vs[i][1] >> c;
		if (c == 'U') 
		{
			vs[i][2] = 0;
			vs[i][3] = 1;
		}
		else if (c == 'D')
		{
			vs[i][2] = 0;
			vs[i][3] = -1;
		}
		else if (c == 'L')
		{
			vs[i][2] = -1;
			vs[i][3] = 0;
		}
		else
		{
			vs[i][2] = 1;
			vs[i][3] = 0;
		}
	}
	TriSearch<double> t;
	double area = 2e18;
	int count = 0;
	double r = t.solve(0, 1e9, [&](double v) {
		double xmin = 1e18, xmax = -1e18, ymin = 1e18, ymax = -1e18;
		for (auto e : vs) {
			double x = e[0] + e[2] * v;
			double y = e[1] + e[3] * v;
			xmin = min(xmin, x);
			xmax = max(xmax, x);
			ymin = min(ymin, y);
			ymax = max(ymax, y);
		}
		auto a = (xmax - xmin) * (ymax - ymin);
		area = min(a, area);
		return a;
		}, [&](double l, double r) {
			count++;
			return count < 150;
		});
	cout << std::fixed << std::setprecision(15) << area << endl;
	return 0;
}

```