小明的飞机快要赶不上了！

幸好大厅的路上有一些传送带。每个传送带都有一定的速度，传送带之间没有重叠。

小明自己行走的速度为w，如果传送带的速度为v的话，在传送带上走的速度就是w+v。

但是小明还是很着急，所以他决定跑一段时间t。他跑的速度是r，那么如果传送带的速度为v的话，在传送带上跑的速度就是r+v。

对于时间t，他不一定要连续跑，可以走走再跑。也不一定非要跑够t。

问小明至少需要多少时间才能到达终点。



输入第一行为用例数T，1<=T<=40。

每一组用例的第一行包含五个整数：

X：为大厅的长度，小明起始位于0，终点是X，1<=X<=1000000

W：为走路的速度

R：为跑步的速度，1<=W<R<=100

t：最多能跑t秒，1<=t<=1000000

n：传送带的个数

接下来的n行，表示n个传送带的详细信息。每行包含三个整数：Bi，Ei，Vi，分别表示传送带的起始位置、终止位置和速度，0<=Bi<Ei<=X，1<=vi<=100

输出包含一个数字，表示至少需要多少时间。输出四舍五入到6位小数。-Xiao Ming aircraft about to catch up !

Fortunately, there are some belt hall way . Each conveyor has a constant speed , there is no overlap between the conveyor .

Bob his walking speed of w, if the belt speed is v, the walking speed of the belt is w+ v.

But Xiao Ming is still very anxious, so he decided to run for some time t. His running speed is r, then if the belt speed of v, the speed of the conveyor belt is running on r+ v.

For the time t, he does not have to run continuously , you can walk rerun . Does not necessarily have to run enough t.

Q. Xiaoming at least the time required to reach the end.



Enter the first act with a number of cases T, 1 <= T <= 40.

Each set of use cases first line contains five integers:

X: is the length of the hall , Xiao Ming starting at 0 , the end is X, 1 <= X <= 1000000

W: the walking speed

R: the running speed , 1 <= W <R <= 100

t: maximum run t seconds , 1 <= t <= 1000000

n: number of conveyor belts

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