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 Python (3.4.3) (judge verdict: Wrong Answer)

import sys
from scipy.optimize import fmin

input = sys.stdin.readline

n = int(input())

XR = []
XL = []
XS = []
YU = []
YD = []
YS = []

for _ in range(n):
    x, y, d = input().split()
    x = int(x)
    y = int(y)
    if d == "D":
        XS.append(x)
        YD.append(y)
    elif d == "U":
        XS.append(x)
        YU.append(y)
    elif d == "R":
        XR.append(x)
        YS.append(y)
    elif d == "L":
        XL.append(x)
        YS.append(y)


max_YD = max(YD + [float("-inf")])
max_YS = max(YS + [float("-inf")])
max_YU = max(YU + [float("-inf")])
min_YD = min(YD + [float("inf")])
min_YS = min(YS + [float("inf")])
min_YU = min(YU + [float("inf")])

max_XR = max(XR + [float("-inf")])
max_XL = max(XL + [float("-inf")])
max_XS = max(XS + [float("-inf")])
min_XR = min(XR + [float("inf")])
min_XL = min(XL + [float("inf")])
min_XS = min(XS + [float("inf")])


def diff_xy(t):
    diff_y = max(max_YD - t, max_YU + t, max_YS) - min(min_YD - t, min_YU + t, min_YS)
    diff_x = max(max_XL - t, max_XR + t, max_XS) - min(min_XL - t, min_XR + t, min_XS)
    return diff_x * diff_y


def binary_search(min_t, max_t):
    mid_t = (min_t + max_t) / 2
    if max_t - min_t < 10 ** (-10):
        return diff_xy(mid_t)

    if diff_xy(min_t) < diff_xy(mid_t):
        return binary_search(min_t, mid_t)
    elif diff_xy(max_t) < diff_xy(mid_t):
        return binary_search(mid_t, max_t)
    else:
        return opt(min_t, mid_t, max_t)


def opt(a, b, c):
    optimal_a = fmin(diff_xy, a, full_output=False, disp=False, xtol= 10 ** (-10))
    optimal_b = fmin(diff_xy, b, full_output=False, disp=False, xtol= 10 ** (-10))
    optimal_c = fmin(diff_xy, c, full_output=False, disp=False, xtol= 10 ** (-10))
    ans = min(diff_xy(optimal_a[0]), diff_xy(optimal_b[0]), diff_xy(optimal_c[0]))
    return(ans)


print(binary_search(0, 2 * 10 ** 8))