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