#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

const int dx[] = { 1, -1, 0, 0 };
const int dy[] = { 0, 0, 1, -1 };

const ld INF = 1e20;

int main()
{
	cout << fixed << setprecision(10);
	int N;
	cin >> N;
	vector<int> x(N), y(N), d(N);
	for (int i = 0; i < N; i++) {
		char c;
		cin >> x[i] >> y[i] >> c;
		d[i] = (c == 'R' ? 0 : c == 'L' ? 1 : c == 'U' ? 2 : 3);
	}
	auto calc = [&](ld t) {
		ld xmi = INF, xma = -INF, ymi = INF, yma = -INF;
		for (int i = 0; i < N; i++) {
			ld tx = x[i] + dx[d[i]] * t;
			ld ty = y[i] + dy[d[i]] * t;
			xmi = min(xmi, tx);
			xma = max(xma, tx);
			ymi = min(ymi, ty);
			yma = max(yma, ty);
		}
		return (xma - xmi) * (yma - ymi);
	};
	ld lb = 0, ub = 1e9;
	int cnt = 120;
	while (cnt--) {
		ld m1 = (lb * 2 + ub) / 3, m2 = (lb + ub * 2) / 3;
		if (calc(m1) < calc(m2)) {
			ub = m2;
		}
		else {
			lb = m1;
		}
	}
	cout << calc(lb) << endl;
	return 0;
}

### Prompt

Generate 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;
using ll = long long;
using ld = long double;

const int dx[] = { 1, -1, 0, 0 };
const int dy[] = { 0, 0, 1, -1 };

const ld INF = 1e20;

int main()
{
	cout << fixed << setprecision(10);
	int N;
	cin >> N;
	vector<int> x(N), y(N), d(N);
	for (int i = 0; i < N; i++) {
		char c;
		cin >> x[i] >> y[i] >> c;
		d[i] = (c == 'R' ? 0 : c == 'L' ? 1 : c == 'U' ? 2 : 3);
	}
	auto calc = [&](ld t) {
		ld xmi = INF, xma = -INF, ymi = INF, yma = -INF;
		for (int i = 0; i < N; i++) {
			ld tx = x[i] + dx[d[i]] * t;
			ld ty = y[i] + dy[d[i]] * t;
			xmi = min(xmi, tx);
			xma = max(xma, tx);
			ymi = min(ymi, ty);
			yma = max(yma, ty);
		}
		return (xma - xmi) * (yma - ymi);
	};
	ld lb = 0, ub = 1e9;
	int cnt = 120;
	while (cnt--) {
		ld m1 = (lb * 2 + ub) / 3, m2 = (lb + ub * 2) / 3;
		if (calc(m1) < calc(m2)) {
			ub = m2;
		}
		else {
			lb = m1;
		}
	}
	cout << calc(lb) << endl;
	return 0;
}

```