Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

2
0 3 D
3 0 L

Sample Output 1

0

After three seconds, the two points will meet at the origin. The value in question will be 0 at that moment.

Sample Input 2

5
-7 -10 U
7 -6 U
-8 7 D
-3 3 D
0 -6 R

Sample Output 2

97.5

The answer may not be an integer.

Sample Input 3

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

Sample Output 3

273

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include"bits/stdc++.h"

using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

#define FOR(k,m,n) for(ll (k)=(m);(k)<(n);(k)++)
#define REP(i,n) FOR((i),0,(n))
#define WAITING(str) int str;std::cin>>str;
#define DEBUGING(str) cout<< #str << " " str<<endl

constexpr int INF = (1 << 30);
constexpr ll INFL = (1ll << 60);
constexpr ll MOD = 1000000007;// 10^9+7
constexpr ld gold = 1.618033988749895;

ld solve(vector<tuple<ll, ll, char>>& v, ld t) {
	ld xmax = -INFL, xmin = INFL, ymax = -INFL, ymin = INFL;
	for (auto& row : v) {
		ld x, y;
		char c;
		tie(x, y, c) = row;
		switch (c) {
		case 'R':x += t; break;
		case 'L':x -= t; break;
		case 'U':y += t; break;
		case 'D':y -= t; break;
		}
		xmax = max(xmax, x);
		xmin = min(xmin, x);
		ymax = max(ymax, y);
		ymin = min(ymin, y);
	}
	return (xmax - xmin)*(ymax - ymin);
}

int main()
{
	ll N;
	cin >> N;
	vector<tuple<ll, ll, char>> v(N);
	REP(i, N) {
		ll x, y;
		char d;
		cin >> x >> y >> d;
		v[i] = make_tuple(x, y, d);
	}

	ld l = 0, r = 1e9;
	ld midl = 0;
	ld midr = 0;

	ld vl = INFL;
	ld vr = INFL;

	while (r - l > 1e-10) {
		if (vl > INFL / 2) {
			midl = (l*gold + r) / (1 + gold);
			vl = solve(v, midl);
		}
		if (vr > INFL / 2) {
			midr = (l + r * gold) / (1 + gold);
			vr = solve(v, midr);
		}

		if (vl < vr) {
			r = midr;
			midr = midl;
			vr = vl;
			vl = INFL;
		}
		else {
			l = midl;
			midl = midr;
			vl = vr;
			vr = INFL;
		}
	}
	cout << fixed << setprecision(10) << solve(v, l) << endl;
	return 0;
}