Elevanagram

It is a well known fact that a number is divisible by 11 if and only if the alternating sum of its digits is equal to 0 modulo 11. For example, 8174958 is a multiple of 11, since 8 - 1 + 7 - 4 + 9 - 5 + 8 = 22.

Given a number that consists of digits from 1-9, can you rearrange the digits to create a number that is divisible by 11?

Since the number might be quite large, you are given integers A1, A2, ..., A9. There are Ai digits i in the number, for all i.

The first line of the input gives the number of test cases, T. T lines follow. Each line contains the nine integers A1, A2, ..., A9.

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is YES if the digits can be rearranged to create a multiple of 11, and NO otherwise.

• 1 ≤ T ≤ 100.
• 1 ≤ A1 + A2 + ... + A9.

• In Sample Case #1, the digits are 336, which can be rearranged to 363. This is a multiple of 11 since 3 - 6 + 3 = 0.
• In Sample Case #2, the digits are 999999999999, which is already a multiple of 11, since 9 - 9 + 9 - 9 + ... - 9 = 0.
• In Sample Case #3, the digits are 5578, which cannot be rearranged to form a multiple of 11.
• In Sample Case #4, the digits are 111234, which can be rearranged to 142131. This is a multiple of 11 since 1 - 4 + 2 - 1 + 3 - 1 = 0.
• In Sample Case #5, the digits are 11177799, which can be rearranged to 19191777. This is a multiple of 11 since 1 - 9 + 1 - 9 + 1 - 7 + 7 - 7 = -22 (which is 0 modulo 11).
• In Sample Case #6, the only digit is 8, which cannot be rearranged to form a multiple of 11.