Problem: Training is indispensable for achieving good results at ICPC. Rabbit wants to win at ICPC, so he decided to practice today as well.

Today's training is to develop physical strength and judgment by running on a straight road. The rabbit is now standing at the starting line and looking out over a long, long road.

There are some carrots along the way, and rabbits can accelerate by eating carrots. Rabbits run at U meters per second when not accelerating, but by eating carrots, the speed is V meters per second from the last carrot eaten to T seconds later. Rabbits can also carry up to K carrots without eating them. Even if you have a carrot, the running speed does not change.

Assuming that it does not take long to bring and eat carrots, I would like to find the shortest time required to reach the goal.



Input


N K T U V L
D1
...
DN


N is the number of carrots, L is the distance from the start to the goal (meters), and Di (1 ≤ i ≤ N) is the distance from the start where the i-th carrot is located (meters).

1 ≤ N ≤ 200, 1 ≤ K ≤ N, 1 ≤ T ≤ 10,000, 1 ≤ U <V ≤ 10,000, 2 ≤ L ≤ 10,000, 0 <D1 <D2 <... <DN <L. All input values ​​are integers.

Output

Output the shortest required time (seconds) on one line. Absolute errors of 10-6 or less are allowed.

Examples

Input

1 1 1 2 3 100
50


Output

49.500000000


Input

3 1 1 2 3 100
49
50
51


Output

48.666666667

Solution (in PYTHON3):

Solution:
from collections import deque
from heapq import *
import sys

sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]

def main():
    n, k, t, u, v, l = MI()
    hp = []
    # (位置、イベントの種類、時間)として、ヒープに突っ込んでいく
    # イベントの種類は　
    # 0 ゴールする
    # 1 ニンジン効果終わる（+食べる）
    # 2 ニンジン拾う（+食べる）　
    # 時間が必要なのは1番だけで他のイベントの時は-1を突っ込んである
    heappush(hp, (l, 0, -1))
    for _ in range(n):
        d = int(input())
        heappush(hp, (d, 2, -1))
    # 「最後に速さが変わった時間、位置」と「ニンジンの数」
    time = pos = caro_n = 0
    # ニンジン効果中ならTrue
    boost = False
    cancel=deque()
    while hp:
        here, e, now = heappop(hp)
        if e == 2:
            if boost:
                if caro_n==k:
                    cancel.append(pos+v*t)
                    time += (here - pos) / v
                    pos=here
                    heappush(hp, (here + v * t, 1, time + t))
                else:
                    caro_n += 1
            else:
                time += (here - pos) / u
                pos = here
                boost = True
                heappush(hp, (here + v * t, 1, time + t))
        if e == 1:
            if cancel and here==cancel[0]:
                cancel.popleft()
                continue
            if caro_n:
                pos=here
                time=now
                caro_n -= 1
                heappush(hp, (here + v * t, 1, now + t))
            else:
                time = now
                pos = here
                boost = False
        if e == 0:
            uv = v if boost else u
            print(time + (l - pos) / uv)
            exit()

main()