# Python 3
 
# Given a 2D matrix of M x N size, find all elements present in every row.
testcase = [
    [7, 1, 3, 5, 3, 6],
    [2, 3, 6, 1, 1, 6],
    [6, 1, 7, 2, 2, 4],
    [6, 6, 7, 1, 3, 3],
    [5, 5, 6, 1, 5, 4],
    [3, 5, 6, 2, 7, 1],
    [4, 1, 4, 3, 6, 4],
    [4, 6, 1, 7, 4, 3]]
# We always need to check elements, so we will use this list to store them.
checked = []
# We will use the elements in the first row one by one to check the other rows.
for elmt in testcase[0]:
    # If this elmt is a duplicate, we skip it.
    if elmt in checked:
        continue
    # If it isn't a duplicate, then add it to the list of checked elmts
    else:
        checked.append(elmt)
    # Reset the counter used to confirm the presence of our elmt.
    count = 1
    # For each row,
    for row in testcase[1:]:
        # Count up if our element is present
        if elmt in row:
            count += 1
        # If a row is missing an elmt, we should break to save a little time.
        else:
            break
    #   If our count is the same number as the amount of rows, then that number
    # must be present in all rows.
    if count == len(testcase):
        try:
            answer.append(elmt)
        # If we had no answers before, make a new list to begin keeping them
        except NameError:
            answer = [elmt]
# Print our answer(s)
try:
    print(answer)
# If there is no answer to print, our answer is None.
except NameError:
    print(None)

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