Ski Accidents
Contest: Codeforces Global Round 8
Difficulty rating: 2500
Tags: constructive algorithms, graphs, greedy

Arthur owns a ski resort on a mountain. There are $$$n$$$ landing spots on the mountain numbered from $$$1$$$ to $$$n$$$ from the top to the foot of the mountain. The spots are connected with one-directional ski tracks. All tracks go towards the foot of the mountain, so there are no directed cycles formed by the tracks. There are at most two tracks leaving each spot, but many tracks may enter the same spot.

A skier can start skiing from one spot and stop in another spot if there is a sequence of tracks that lead from the starting spot and end in the ending spot. Unfortunately, recently there were many accidents, because the structure of the resort allows a skier to go through dangerous paths, by reaching high speed and endangering himself and the other customers. Here, a path is called dangerous, if it consists of at least two tracks.

Arthur wants to secure his customers by closing some of the spots in a way that there are no dangerous paths in the resort. When a spot is closed, all tracks entering and leaving that spot become unusable.

Formally, after closing some of the spots, there should not be a path that consists of two or more tracks.

Arthur doesn't want to close too many spots. He will be happy to find any way to close at most $$$\frac{4}{7}n$$$ spots so that the remaining part is safe. Help him find any suitable way to do so.

Input format:
The first line contains a single positive integer $$$T$$$ — the number of test cases. $$$T$$$ test case description follows.

The first line of each description contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$) — the number of landing spots and tracks respectively.

The following $$$m$$$ lines describe the tracks. Each of these lines contains two integers $$$x$$$ and $$$y$$$ ($$$1 \leq x < y \leq n$$$) — indices of the starting and finishing spots for the respective track. It is guaranteed that at most two tracks start at each spot. There may be tracks in which starting and finishing spots both coincide.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output format:
For each test case, print a single integer $$$k$$$ ($$$0 \leq k \leq \frac{4}{7}n$$$) — the number of spots to be closed. In the next line, print $$$k$$$ distinct integers — indices of all spots to be closed, in any order.

If there are several answers, you may output any of them. Note that you don't have to minimize $$$k$$$. It can be shown that a suitable answer always exists.

Constraints:
Time limit: 2.0 s
Memory limit: 512.0 MB

Note:
In the first sample case, closing any two spots is suitable.

In the second sample case, closing only the spot $$$1$$$ is also suitable.

Example input:
2
4 6
1 2
1 3
2 3
2 4
3 4
3 4
7 6
1 2
1 3
2 4
2 5
3 6
3 7

Example output:
2
3 4 
4
4 5 6 7

Editorial:
Let's consider vertices from $$$1$$$ to $$$n$$$ (that is, in topological order). We divide them into three sets $$$V_0$$$, $$$V_1$$$, $$$V_2$$$:
• $$$V_0$$$ contains all vertices that only have incoming edges from $$$V_2$$$;
• $$$V_1$$$ contains all vertices that have an incoming edge from $$$V_0$$$, but not from $$$V_1$$$;
• $$$V_2$$$ contains all vertices that have an incoming edge from $$$V_1$$$.
It is not hard to see that erasing all vertices in $$$V_2$$$ leaves no two-edge paths. The same solution can be simply rephrased as: go from left to right, and remove current vertex if it is at the end of a two-edge path.
Why does this work? Every vertex of $$$V_2$$$ has to have at least one incoming edge from $$$V_1$$$. There are at most $$$2|V_1|$$$ such edges, thus $$$|V_2| \leq 2|V_1|$$$, and $$$|V_1| \geq |V_2| / 2$$$. Similarly, $$$|V_0| \geq |V_1| / 2 \geq |V_2| / 4$$$. But then $$$n = |V_0| + |V_1| + |V_2| \geq |V_2| (1 + 1/2 + 1/4) = 7|V_2|/4$$$, thus $$$|V_2| \leq 4n/7$$$.
This is very easy to implement in $$$O(n)$$$ time.