PHP function to check a site position in Google Search Results for an exact phrase

$qst is a word (phrase) to search in Google Search Results (test word) $qsd is a web page url address (https://*.[login to view URL]*) $num is the amount of results to fetch in SERP We need you to write the function google_pos($qst , $qsd, $num=100) { must return a result array with elements ['0'] - the number of results in SERP per given word when rhe result is valid or on error -1 invalid word - if strlen($qst) <3 and >64 -2 invalid url syntax in $qsd -3 invalid url lenght strlen($qsd <6 and >96) -4 google do not provide results i.e. if the web server we are checking from is blocked due to many requests in a short time ['gPosOrg'] - the position where the DOMAIN name is FOUND on the organic search results ($num results per page) if not found- 0 if found from 1 to $num ['exact'] = 1 if the result matches the full url, 0 if only a domain match ['adwords'] - the amount of results in Google Ads (=0 if no Google paid ads results) We will use the script in [login to view URL] mainly with cirilic results example search url is $google_url = '[login to view URL]'.str_replace(" ", "%20", $qst).'&num='.$num; an additional parameter for the page encoiding must be provided as we want to use the script in both cp1251 and utf8 encoding. Provide the best available ini_set('user_agent',' ... and note the script must work as naturaly as possible as we need to check a big amount of words