human
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 4/7n spots so that the remaining part is safe. Help him find any suitable way to do so.

Input

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 ≤ n ≤ 2 ⋅ 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 ≤ x < y ≤ 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 ⋅ 10^5.

Output

For each test case, print a single integer k...
Solve the task in PYTHON3.

What is a Python code to find any way to close at most 4/7n spots so that the remaining part is safe?
assistant
#Here is Python code to find any way to close at most 4/7n spots so that the remaining part is safe.

#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading 
from math import floor
# from bisect import bisect_right
from collections import Counter
from math import gcd

mod=998244353


def main():
    for _ in range(int(input())):
        n,m=map(int,input().split())
        graph=[[] for i in range(n)]

        for i in range(m):
            u,v=map(lambda x:int (x)-1,input().split())
            graph[v].append(u)

        # print(graph)
        delete=[False]*n
        for i in range(n):
            # print(delete)
            for p1 in graph[i]:
                for p2 in graph[p1]:
                    if not (delete[p1] or delete[p2]):
                        delete[i]=True

        ans=[(i+1) for i in range(n) if delete[i]]
        print(len(ans))
        print(*ans)

# 5
# 991 143 445 903 399


BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def...