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 pb          emplace_back
#define ll          long long
#define fi          first
#define se          second
#define mp          make_pair

using namespace std;

typedef long double ld;
typedef tuple<ld, ld, char> TPoint;
const int N = int(1e5) + 2;
const int nTest = 300;

TPoint p[N];
int _x, _y, n;
char d;
ld x, y;

ld f(ld step) {
    ld xmax = -1e18, xmin = 1e18, ymax = -1e18, ymin = 1e18;
    for(int i = 1; i <= n; ++i) {
        tie(x, y, d) = p[i];
        if(d == 'U') y += step;
        else if(d == 'D') y -= step;
        else if(d == 'L') x -= step;
        else x += step;
        xmax = max(xmax, x), xmin = min(xmin, x);
        ymax = max(ymax, y), ymin = min(ymin, y);
    }
    return (xmax - xmin) * (ymax - ymin);
}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    #define Task        "test"
    if(fopen(Task".inp", "r")) {
        freopen(Task".inp", "r", stdin);
        freopen(Task".out", "w", stdout);
    }
    cin >> n;
    for(int i = 1; i <= n; ++i) {
        cin >> _x >> _y >> d;
        p[i] = make_tuple(_x, _y, d);
    }
    ld lef = 0, rig = 2e12, res = 1e18;
    for(int i = 0; i < nTest; ++i) {
        ld x1 = lef + (rig - lef) / 3.0;
        ld x2 = rig - (rig - lef) / 3.0;
        if(f(x1) > f(x2)) lef = x1;
        else rig = x2;
    }
    cout << fixed << setprecision(10) << f(rig);
}