Haskell function that returns a list of elements in a list with more than given amount of occurrences

I tried making a function that as in the title takes 2 arguments, a number that specifies how many times the number must occur and a list that we are working on, I made a function that counts number of appearances of given number in a list and I tried using it in my main function, but I cannot comprehend how the if else and indentations work in Haskell, it's so much harder fixing errors than in other languages, i think that I'm missing else statement but even so I don't know that to put in there

count el list = count el list 0
     where count el list output
             | list==[] = output
             | head(list)==el = count el (tail(list)) output+1
             | otherwise = count el (tail(list)) output


moreThan :: Eq a => Int -> [a] -> [a]
moreThan a [] = []
moreThan a list = moreThan a list output i
    where moreThan a list [] 0
            if i == length (list)
                then output
            else if elem (list!!i) output
                 then moreThan a list output i+1
            else if (count (list!!i) list) >= a 
                then moreThan a list (output ++ [list!!i]) i+1 

All I get right now is

parse error (possibly incorrect indentation or mismatched brackets)