Andy has no time for ores, he's just here for the ANDs.

Andy has (N) pairs of integers, with each integer given in binary as a bitstring. The (i)th pair consists of bitstrings (A_i) and (B_i). No bitstring has leading zeros (except for the bitstring "0" itself).

Andy may choose at most (K) of these pairs for swapping. For each chosen pair (i), he'll swap its bitstrings (A_i) and (B_i).

After all of these swaps, he'll compute (X) as the bitwise AND of the (N) bitstrings (A_{1..N}), and (Y) as the bitwise AND of the (N) bitstrings (B_{1..N}), and finally add (X) and (Y) together. What's the maximum sum of (X + Y) that Andy can achieve?

Note that:

The integer value of a bitstring (S) is equal to (S_1 * 2^{|S|-1} + S_2 * 2^{|S|-2} + ... + S_{|S|} * 2^0).
The bitwise AND of a set of bitstrings (S) is a bitstring (V) such that, if leading zeros were appended to bitstrings in (S) until they all had the same length, then (V) also has that same length, with (V_i) equal to 1 if and only if all bitstrings in (S) have a 1 at index (i).

Constraints

(1 \le T \le 90) (1 \le N \le 2{,}000{,}000) (0 \le K \le N) (0 \le A_{i,j}, B_{i,j} \le 1)

The total length of all bitstrings (A_{1..N}) and (B_{1..N}) is at most (4{,}000{,}000). The total length of all bitstrings across all test cases is at most (10{,}000{,}000).

Input

Input begins with an integer (T), the number of test cases. For each test case, there is first a line containing (2) space-separated integers, (N) and (K). Then, (N) lines follow, the (i)th of which contains (2) space-separated bitstrings, (A_i) and (B_i).

Output

For the (i)th test case, print a line containing "Case #i: " followed by a single bitstring, the maximum achievable sum of bitstring ANDs, with no leading zeros (unless the bitstring is "0").

Sample Explanation

In the first case, (X = 1011011_2 ,(91_{10})), (Y = 101101_2 ,(45_{10})), and (X + Y = 10001000_2 ,(136_{10})).

In the second case, no swaps are allowed, so we must have (X = 1011011_2) AND (0_2 = 0_2) and (Y = 0_2 ) AND (101101_2 = 0_2).

In the third case, one option is to swap (A_1) and (B_1). This yields (X = 0_2) AND (0_2 = 0_2) and (Y = 1011011_2) AND (101101_2 = 1001_2).

In the fourth case, one option is to swap (A_1) and (B_1) as well as (A_3) and (B_3), resulting in (X = 111_2 ,(7_{10})), (Y = 1_2 ,(1_{10})), and (X + Y = 1000_2 ,(8_{10})).