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

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <ctime>
#include <bitset>
#include <complex>

using namespace std;

typedef long double ld;

const ld INF = 1e18 + 239;

int n;

vector<int> x;
vector<int> y;
vector<int> dir;

ld solve(ld t) {
	ld xf = INF;
	ld xs = -INF;
	ld yf = INF;
	ld ys = -INF;
	for (int i = 0; i < n; i++) {
		ld tx = x[i];
		ld ty = y[i];
		if (dir[i] == 0) tx += t;
		if (dir[i] == 1) tx -= t;
		if (dir[i] == 2) ty += t;
		if (dir[i] == 3) ty -= t;
		xf = min(xf, tx);
		xs = max(xs, tx);
		yf = min(yf, ty);
		ys = max(ys, ty);
	}
	return (xs - xf) * (ys - yf);
}

int xmin(int t) {
	int ans = INF;
	for (int i = 0; i < n; i++) {
		ans = min(ans, x[i] + t * (dir[i] == 'R') - t * (dir[i] == 'L'));
	}
	return ans;
}

int xmax(int t) {
	int ans = INF;
	for (int i = 0; i < n; i++) {
		ans = min(ans, x[i] + t * (dir[i] == 'R') - t * (dir[i] == 'L'));
	}
	return ans;
}



signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	cout << fixed << setprecision(16);

	cin >> n;
	x.resize(n);
	y.resize(n);
	dir.resize(n);
	for (int i = 0; i < n; i++) {
		cin >> x[i] >> y[i];
		char c;
		cin >> c;
		if (c == 'R') dir[i] = 0;
		if (c == 'L') dir[i] = 1;
		if (c == 'U') dir[i] = 2;
		if (c == 'D') dir[i] = 3;
	}
	ld l = 0;
	ld r = 1e18 + 239;
	for (int i = 0; i < 500; i++) {
		ld tl = l + (r - l) / 3;
		ld tr = r - (r - l) / 3;
		if (solve(tl) > solve(tr)) {
			l = tl;
		} else {
			r = tr;
		}
	}
	cout << solve(r) << endl;
}