For non-negative integers (x, y) and a positive integer (k), let (S(x, y, k)) be the set of values ((x + i) \oplus (y + j)) for all (0 \le i, j < k). Formally: () S(x, y, k) = \{ (x + i) \oplus (y + j) \mid 0 \le i, j < k \} () where (\oplus) denotes the bitwise XOR operation. Define (f(x, k)) as the maximum size of (S(x, y, k)) over all non-negative integers (y) (that is, (y \ge 0)). You are given integers (x) and (k). Compute (f(x, k)). Each test contains multiple test cases. The first line contains the number of test cases (t) ((1 \le t \le 10^4)). The description of the test cases follows. Each of the next (t) lines contains two integers (x) and (k) ((1 \le x, k \le 10^{17})). For each test case, output a single integer — the value of (f(x, k)). In the first example, since (k=1), the set (S) will always contain exactly one element regardless of (y). For instance, if we pick (y = 69), we have (S(67, 69, 1) = \{67 \oplus 69\} = \{6\}), so (|S(x, y, k)| = 1). In the second example, we have (x = 7) and (k = 3). The optimal choice is (y = 8). The values of ((x + i) \oplus (y + j)) are: () 7 \oplus 8, 7 \oplus 9, 7 \oplus 10, 8 \oplus 8, 8 \oplus 9, 8 \oplus 10, 9 \oplus 8, 9 \oplus 9, 9 \oplus 10 () which simplifies to (15, 14, 13, 0, 1, 2, 1, 0, 3). The set of distinct values is (S(7, 8, 3) = \{0, 1, 2, 3, 13, 14, 15\}), so the size is (7). It can be shown that no other (y) yields a larger size. However, the choice of (y) matters; for example, if you chose (y = 22), you would get (S(7, 22, 3) = \{16, 17, 30, 31\}) with size (4), which is suboptimal. In the sixth example, after countless calculations, we managed to figure out that the optimal (y) is (278\,302\,368\,699\,121\,665), which gives an answer of (398\,158\,383\,604\,301\,822). The proof is left to the reader as a tri