Write a function that accepts a string argument and returns a link representing the mp4 file of the most popular coub

[odv]

API: https://coub.com/dev/docs/Coub+API%2FOverview Input: string representing a search query, for example: 'best coub ever'

Please sort by 'newest_popular'.

Output example: https://coubsecure-s.akamaihd.net/get/b122/p/coub/simple/cw_video_for_sharing/690b14e0d62/faf6703df0169864806fc/1568905880_looped_1568905875.mp4

Please note that there is a hint in the URL. If the API returns "none" please show this link instead: https://www.youtube.com/watch?v=CYfjFb7WjGQ

[Deobp]

fetch('https://coub.com/api/v2/search?q=best%20coub%20ever&order_by=newest_popular')
    .then((response) => {
        return response.json();
    })
    .then((data) => {
        return data.coubs[0].file_versions.share.default
    })
    .catch(error => ('https://www.youtube.com/watch?v=CYfjFb7WjGQ'));