Clarify the Libre license of website contents :)

I'm trying to publish more OrganicMaps screenshots on Wikimedia Commons but I'm not 100% sure about the license of our screenshots. Example:

https://commons.wikimedia.org/wiki/File:Organic-map--screenshot--dark.jpg

Maybe we clarify this license from the official website itself :) At the moment this is "somehow implicit" from the repository that says Apache 2.0 (but maybe that should cover just source code, not contents?).

For example, maybe in the website we can put a small phrase like this one:

<p xmlns:cc="http://creativecommons.org/ns#" xmlns:dct="http://purl.org/dc/terms/"><span property="dct:title">OrganicMaps</span> is licensed under <a href="http://creativecommons.org/licenses/by-sa/4.0/?ref=chooser-v1" target="_blank" rel="license noopener noreferrer" style="display:inline-block;">CC BY-SA 4.0<img style="height:22px!important;margin-left:3px;vertical-align:text-bottom;" src="https://mirrors.creativecommons.org/presskit/icons/cc.svg?ref=chooser-v1"><img style="height:22px!important;margin-left:3px;vertical-align:text-bottom;" src="https://mirrors.creativecommons.org/presskit/icons/by.svg?ref=chooser-v1"><img style="height:22px!important;margin-left:3px;vertical-align:text-bottom;" src="https://mirrors.creativecommons.org/presskit/icons/sa.svg?ref=chooser-v1"></a></p> 

Or something like that :) I generated that from here:

https://chooser-beta.creativecommons.org/

Thanks for any opinion :)

Review the website source and the repository's current Apache 2.0 licensing information, then determine which license applies to website contents and screenshots. Done means the intended license is agreed and an explicit notice is added to the official website.