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 PYTHON3):

Solution:
n = int(input())
xyd = [input().split() for _ in range(n)]


x_max_RL_ = [-2e8] * 3
x_min_RL_ = [2e8] * 3
y_max_UD_ = [-2e8] * 3
y_min_UD_ = [2e8] * 3

for x, y, d in xyd:
    x = int(x)
    y = int(y)

    if d == "R":
        ho, ve = 2, 1
    elif d == "L":
        ho, ve = 0, 1
    elif d == "U":
        ho, ve = 1, 2
    else:  # d == "D"
        ho, ve = 1, 0
    x_max_RL_[ho] = max(x_max_RL_[ho], x)
    x_min_RL_[ho] = min(x_min_RL_[ho], x)
    y_max_UD_[ve] = max(y_max_UD_[ve], y)
    y_min_UD_[ve] = min(y_min_UD_[ve], y)

kouho_time = set([0])

for _max_min in [x_max_RL_, x_min_RL_, y_max_UD_, y_min_UD_]:
    for i in range(0, 3):
        if abs(_max_min[i]) <= 1e8:
            for j in range(i + 1, 3):
                if abs(_max_min[j]) <= 1e8:
                    cross = (_max_min[i] - _max_min[j]) / (j - i)
                    if cross > 0:
                        kouho_time.add(cross)


ans = 4e16
for t in kouho_time:
    x_max = max(
        x_max_RL_[i] + t * (i - 1) for i in range(3) if abs(x_max_RL_[i]) <= 1e8
    )
    x_min = min(
        x_min_RL_[i] + t * (i - 1) for i in range(3) if abs(x_min_RL_[i]) <= 1e8
    )
    y_max = max(
        y_max_UD_[i] + t * (i - 1) for i in range(3) if abs(y_max_UD_[i]) <= 1e8
    )
    y_min = min(
        y_min_UD_[i] + t * (i - 1) for i in range(3) if abs(y_min_UD_[i]) <= 1e8
    )

    ans = min(ans, (x_max - x_min) * (y_max - y_min))

print(ans)