Trying to solve this problem below Compound words, unable to find a logic to return all variants of a word in python.. any help would be appreciated

Question

Compound words problem, Here's the code I have tried below, but this doesn't work, it won't return all the possible pairs, it only returns one pair of a word ex:

"verylongjump": ["very", "long", "jump"])

but it will not return both:

"verylongjump": ["very", "long", "jump"]   
"verylongjump": ["very","longjump"]

Question Input:

Input:

["long", "very", "jump", "verylongjump", "longjump", 
"iconstar", "star", "icon", "icons", "tar", "stars,"iconstars"]

Expected Output:

"verylongjump": ["very", "long", "jump"]
"verylongjump": ["very","longjump"]
"iconstar": ["icon",""star"]
"iconstar":["icons", "tar"]
"iconstars":["icon","stars"]

def concatenatedWords(words):
    # Intitialize empty array for result set
    result = []

    # Helper function for Work Break II without DP but accepted by LeetCode
    def helper(idx, word, wordSet, curr, result):
        if idx == len(word):
            result.append(curr)
            return
        if idx > len(word):
            return
        for i in range(idx+1, len(word)+1):
            if word[idx:i] in wordSet:
                helper(i, word, wordSet, curr + [word[idx:I]], result)
        return

    # Convert the list of words into a HashSet
    wordSet = set(words)
    
    # Find word breaks for each word
    for word in words:
        # Initialize current word split result set
        curr = []
        helper(0, word, wordSet, [], curr)

        # Word Break solution captures the word themselves as they are present in the 
        # hash set; Filter them out by checking for length of current splits more than 1
        if len(curr) > 1:
            res = []
            # Flatten the result list
            for elem in curr:
                res.extend(elem)
        result.append(res)
    return result