Strange Radiation
Contest: Codeforces Round 425 (Div. 2)
Difficulty rating: 2500
Tags: binary search, implementation, math

n people are standing on a coordinate axis in points with positive integer coordinates strictly less than 106. For each person we know in which direction (left or right) he is facing, and his maximum speed.

You can put a bomb in some point with non-negative integer coordinate, and blow it up. At this moment all people will start running with their maximum speed in the direction they are facing. Also, two strange rays will start propagating from the bomb with speed s: one to the right, and one to the left. Of course, the speed s is strictly greater than people's maximum speed.

The rays are strange because if at any moment the position and the direction of movement of some ray and some person coincide, then the speed of the person immediately increases by the speed of the ray.

You need to place the bomb is such a point that the minimum time moment in which there is a person that has run through point 0, and there is a person that has run through point 106, is as small as possible. In other words, find the minimum time moment t such that there is a point you can place the bomb to so that at time moment t some person has run through 0, and some person has run through point 106.

Input format:
The first line contains two integers n and s (2 ≤ n ≤ 105, 2 ≤ s ≤ 106) — the number of people and the rays' speed.

The next n lines contain the description of people. The i-th of these lines contains three integers xi, vi and ti (0 < xi < 106, 1 ≤ vi < s, 1 ≤ ti ≤ 2) — the coordinate of the i-th person on the line, his maximum speed and the direction he will run to (1 is to the left, i.e. in the direction of coordinate decrease, 2 is to the right, i.e. in the direction of coordinate increase), respectively.

It is guaranteed that the points 0 and 106 will be reached independently of the bomb's position.

Output format:
Print the minimum time needed for both points 0 and 106 to be reached.

Your answer is considered correct if its absolute or relative error doesn't exceed 10 - 6. Namely, if your answer is a, and the jury's answer is b, then your answer is accepted, if $${ \frac { | a - b | } { \operatorname* { m a x } ( 1, | b | ) } } \leq 1 0 ^ { - 6 }$$.

Constraints:
Time limit: 3.0 s
Memory limit: 256.0 MB

Note:
In the first example, it is optimal to place the bomb at a point with a coordinate of 400000. Then at time 0, the speed of the first person becomes 1000 and he reaches the point 106 at the time 600. The bomb will not affect on the second person, and he will reach the 0 point at the time 500000.

In the second example, it is optimal to place the bomb at the point 500000. The rays will catch up with both people at the time 200. At this time moment, the first is at the point with a coordinate of 300000, and the second is at the point with a coordinate of 700000. Their speed will become 1500 and at the time 400 they will simultaneously run through points 0 and 106.

Example input:
2 999
400000 1 2
500000 1 1

Example output:
500000.000000000000000000000000000000

Example input:
2 1000
400000 500 1
600000 500 2

Example output:
400.000000000000000000000000000000

Editorial:
We'll use binary search by answer. Obviously, that answer is always less than 106 and more than 0. We denote the answer for t at current iteration.
For each person, running left, we have to find positions of bomb, at which he will have time to reach the point 0 in time t. Let d be the distance between the person and point 0 and d1 be the distance, which passed the person before he caught up with the ray.
If $$\frac{d}{v_i} \leq t$$, we can place bomb in every point.
Otherwise, we can place bomb in point x, $$\begin{cases}
\frac{d_1}{v_i} + \frac{d - d_1}{v_i + s} \leq t, \\
x \geq d, \\
x = (d - d_1) + \frac{d_1 \cdot s}{v_i}.
\end{cases}$$
1. Before the meeting with the ray, person ran a distance of d1 at a speed of vi, after the meeting he ran a distance of d - d1 at a speed of vi + s. We require the total time be no more than t.
2. We need the person to be caught by rays, so bomb have to have coordinate more than person's initial coordinate.
3. We know that rays and person met at the point of d - d1 at the moment of $$\frac{d_{1}}{v_{i}}$$. Rays moves at the speed of s. It means they started moving at the point of $$( d - d _ { 1 } ) + \frac { d _ { 1 } \cdot s } { v _ { i } }$$.
Note that solutions of this system form a segment on the coordinate line.
For persons, running right, the reasoning is similar.
We find all the segments for persons, runnig left, and persons, running right. If some point with the whole coordinate belongs to segment for person, running left and person, running right, we move right border of binary search and left border otherwise.
To find this point we can use scanline or prefix sums. Let the binary search make it iterations. Than final asymptotics is O(n · log n · it), if we use scanline, or O((n + maxx) · it) if we use prefix sums.