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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
const int inf = 1e9 + 7;
const ll INF = 1e18;
const int mod = 1000000007;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<typename T> T pow(T a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }
template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { for (auto &vi: vec) os << vi << " "; return os; }
template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<typename T >inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T> inline void add(T &a, int b) { a += b; if (a >= mod) a -= mod; }


void solve();

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(10);

    int T;
    // cin >> T;
    T = 1;

    while (T--) {
        solve();
    }
}

void solve() {
  clock_t ST, ED;
  ST = clock();

  int n;
  cin >> n;
  vector<double> X(n), Y(n);
  vector<char> D(n);
  rep(i, n) cin >> X[i] >> Y[i] >> D[i];

  double lma = -inf, lmi = inf, rma = -inf, rmi = inf, udma = -inf, udmi = inf;
  rep(i, n) {
    if (D[i] == 'L') { chmax(lma, X[i]); chmin(lmi, X[i]); }
    else if (D[i] == 'R') { chmax(rma, X[i]); chmin(rmi, X[i]); }
    else { chmax(udma, X[i]); chmin(udmi, X[i]); }
  }
  // xmax - xmin
  auto calcx = [&](double t) {
    double xmin = min({udmi, lmi - t, rmi + t});
    double xmax = max({udma, lma - t, rma + t});
    return xmax - xmin;
  };

  double uma = -inf, umi = inf, dma = -inf, dmi = inf, lrma = -inf, lrmi = inf;
  rep(i, n) {
    if (D[i] == 'U') { chmax(uma, Y[i]); chmin(umi, Y[i]); }
    else if (D[i] == 'D') { chmax(dma, Y[i]); chmin(dmi, Y[i]); }
    else { chmax(lrma, Y[i]); chmin(lrmi, Y[i]); }
  }
  // cout << uma << " " << umi << " " << dma << " " << dmi << " " << lrma << " " << lrmi << endl;
  auto calcy = [&](double t) {
    double ymin = min({lrmi, dmi - t, umi + t});
    double ymax = max({lrma, dma - t, uma + t});
    return ymax - ymin;
  };

  auto f = [&](double t) {
    return calcx(t) * calcy(t);
  };
  double l = 0, r = inf;
  double eps = 1e-10;
  while (abs(l - r) > eps) {
    double l2 = (l * 2 + r) / 3;
    double r2 = (l + r * 2) / 3;
    if (f(l2) < f(r2)) r = r2;
    else l = l2;

    ED = clock();
    if (1.0 * (ED - ST) / CLOCKS_PER_SEC > 1.5) break;
  }
  cout << f(l) << endl;
}