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:
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
using namespace std;
#define FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define MT make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
#define RT return
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;

void solve() {
    ll ma[2][3], mi[2][3];
    rep(i, 2)rep(j, 3) {
        ma[i][j] = LLONG_MIN / 3;
        mi[i][j] = LLONG_MAX / 3;
    }

    auto put = [&](int i, int j, ll z) {
        smax(ma[i][j], z);
        smin(mi[i][j], z);
    };

    int N;
    cin >> N;
    rep(i, N) {
        ll x, y;
        char c;
        cin >> x >> y >> c;
        x *= 2;
        y *= 2;
        if (c == 'R') {
            put(0, 0, x);
            put(1, 2, y);
        } else if (c == 'L') {
            put(0, 1, x);
            put(1, 2, y);
        } else if (c == 'U') {
            put(0, 2, x);
            put(1, 0, y);
        } else {
            put(0, 2, x);
            put(1, 1, y);
        }
    }

    ll ans = LLONG_MAX;
    rep(t, 100000001) {
        static ll A[2];
        rep(i, 2) {
            ll large = max({ ma[i][0] + t, ma[i][1] - t, ma[i][2] });
            ll small = min({ mi[i][0] + t, mi[i][1] - t, mi[i][2] });
            A[i] = large - small;
        }
        smin(ans, A[0] * A[1]);
    }
    
    cout << ans / 4.0 << endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout << fixed << setprecision(15);
	solve();
	return 0;
}