'''
Algorithm:

Or & and all bit strings together.
Where the digits do not match, there should be a '*' in the final result,
Otherwise, the corresponding digit in the result should be the same digit.

If 2 ^ the number of differences > than the number of bit-strings inputted,
there are no results, otherwise output the found result.
'''

def main(index):
    # Input bit-strings, ignore first one, ensure that all are distinct
    input = list(set(raw_input(str(index) + ". ").split(", ")[1:]))
    if(len(input) == 0): return
    # End results for operations
    super_and, super_or = (['1' for i in range(len(input[0]))],
                           ['0' for i in range(len(input[0]))])
    # The count of the number of differences
    odd_count = 0
    # Iterate over digits
    for i in range(len(input[0])):
        # Iterate over inputs
        for i2 in range(len(input)):
            # And it to the rolling and
            super_and[i] = str(int(super_and[i]) & int(input[i2][i]))
            # Or it to the rolling or
            super_or[i] = str(int(super_or[i]) | int(input[i2][i]))
        # If there is a diff. it is '*'
        if not super_or[i] == super_and[i]: super_and[i] = "*"; odd_count += 1
    # No combination spans output
    if 2**odd_count > len(input): print "NONE"
    # Print result
    else: print str(index) + ". " + "".join(super_and)

if __name__ == "__main__":
    for i in range(5):
        failed = False
        while True:
            try:
                main(i + 1)
                break
            except ValueError:
                print "Invalid Input"