The bitwise nor(^{\text{∗}}) of an array of (k)-bit integers (b_1, b_2, \ldots, b_m) can be computed by calculating the bitwise nor cumulatively from left to right. More formally, (\operatorname{nor}(b_1, b_2, \ldots, b_m) = \operatorname{nor}(\operatorname{nor}(b_1, b_2, \ldots, b_{m - 1}), b_m)) for (m\ge 2), and (\operatorname{nor}(b_1) = b_1). You are given an array of (k)-bit integers (a_1, a_2, \ldots, a_n). For each index (i) ((1\le i\le n)), find the maximum bitwise nor among all subarrays(^{\text{†}}) of (a) containing index (i). In other words, for each index (i), find the maximum value of (\operatorname{nor}(a_l, a_{l+1}, \ldots, a_r)) among all (1 \le l \le i \le r \le n). (^{\text{∗}}) The logical nor of two boolean values is (1) if both values are (0), and (0) otherwise. The bitwise nor of two (k)-bit integers is calculated by performing the logical nor operation on each pair of the corresponding bits. For example, let us compute (\operatorname{nor}(2, 6)) when they are represented as (4)-bit numbers. In binary, (2)=(0010_2) and (6=0110_2). Therefore, (\operatorname{nor}(2,6) = 1001_2 = 9) as by performing the logical nor operations from left to right, we have: (\operatorname{nor}(0,0) = 1) (\operatorname{nor}(0,1) = 0) (\operatorname{nor}(1,0) = 0) (\operatorname{nor}(1,1) = 0) Note that if (2) and (6) were represented as (3)-bit integers instead, then (\operatorname{nor}(2,6) = 1). (^{\text{†}})An array (x) is a subarray of an array (y) if (x) can be obtained from (y) by the deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. 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. The first line of