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Predict the correct output of the below code snippet.
sc.parallelize(["C,29","B,9","A,18"]).map(lambda x:x.split(",")).sortBy(lambda x:x[1]).collect()
[['B', '9'], ['A', '18'], ['C', '29']]
[['A', '18'], ['B', '9'], ['C', '29']]
[['B', 9], ['A', 18], ['C', 29]]
[['A', '18'], ['C', '29'], ['B', '9']]
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