Q: I have a few questions that I am hoping you can answer: 1.
Does Curatora support automated posting to Instagram? For example, the ability to schedule a post for a specific date and time, and it posts automatically to the desired Instagram account.
2. Does Curatora support Instagram and Twitter RSS feeds?
3. How does Curatora compare to something like RSS Ground or RSS.com?
4. What kind of runway does your team currently have for Curatora specifically? I ask since you mentioned in another response that this is one of other studio projects, and it's a cost-effective model. A rough ballpark is fine.
Thanks in advance for your time and responces.

Imtiyaz_Curatora
May 14, 2024A: Hello Andregivenchy,
Thank you for your interest in Curatora! I'd be happy to answer your questions.
1. Yes, Curatora does support automated posting to Instagram. You can schedule your posts in advance and Curatora will automatically post them to your desired Instagram account at the scheduled time.
2. Curatora supports custom RSS Feed / Sources. You can easily add them to your Content Streams and use them to curate content for your social media accounts. But as of now, you cannot add direct Twitter or Instagram Feed URLs but a valid RSS Feed URL.
3. Curatora is a content curation platform that focuses on helping businesses find and share engaging content on social media. RSS Ground and RSS.com are both RSS feed management tools that allow you to collect and organize RSS feeds from various sources. While all three tools have similar features, Curatora offers a more robust set of features for social media content curation, including automatic scheduling of posts, content categorization, and hashtag recommendations.
4. Yes, Curatora is a startup product developed by Sizmic Labs, which is a startup studio. Sizmic Labs has sufficient resources to manage all of its products, including Curatora, and has dedicated short teams and shared teams for each product.
Please feel free to reach me at imtiyaz@curatora.io, if you have any questions. I'd happy to answer.