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 <bits/stdc++.h>

#define F first
#define S second
#define pb push_back
#define ll long long
#define all(v) (v).begin(),(v).end()
#define endl '\n'

using namespace std;

const int N = 100005;
const int MOD = 1e9 + 7;
const int INF = 2000000000;
const ll INFLL = 1LL * INF * INF;

#define ld long double

int n, x[N], y[N];
char d[N];

ld get(ld k) {
    ld mxx = -INF, mnx = INF, mxy = -INF, mny = INF;
    for (int i = 1; i <= n; i++) {
        ld X = x[i], Y = y[i];
        if (d[i] == 'U') Y += k;
        if (d[i] == 'D') Y -= k;
        if (d[i] == 'R') X += k;
        if (d[i] == 'L') X -= k;
        mxx = max(mxx, X); mnx = min(mnx, X);
        mxy = max(mxy, Y); mny = min(mny, Y);
    }
    return (mxx - mnx) * (mxy - mny);
}

int main() {
    ios_base :: sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> x[i] >> y[i] >> d[i];
    }
    int mxx = x[1], mnx = x[1], mxy = y[1], mny = y[1];
    vector <int> up, down, left, right;
    for (int i = 1; i <= n; i++) {
        if (d[i] == 'U') up.pb(i);
        if (d[i] == 'D') down.pb(i);
        if (d[i] == 'L') left.pb(i);
        if (d[i] == 'R') right.pb(i);
        mxx = max(mxx, x[i]); mnx = min(mnx, x[i]);
        mxy = max(mxy, y[i]); mny = min(mny, y[i]);
    }
    ld l = 0, r = 1e9, res = INFLL;
    for (int go = 0; go < 600; go++) {
        ld m1 = l + (r - l) / 3.0, m2 = r - (r - l) / 3.0;
        ld t1 = get(m1), t2 = get(m2);
        if (t1 < t2) {
            res = min(res, t1);
            r = m2;
        } else {
            res = min(res, t2);
            l = m1;
        }
    }
    cout << fixed << setprecision(16) << res << endl;
    return 0;
}