a=[*map(int,[*open(0)][1].split())]
for k in 0,1:
for i in range(19):
z=1<<i
for j in range(len(a)):
if j&z:a[j-k*z]^=a[j+k*z-z]
print(*reversed(a))
1920. Build Array from Permutation | 494. Target Sum |
797. All Paths From Source to Target | 1547B - Alphabetical Strings |
1550A - Find The Array | 118B - Present from Lena |
27A - Next Test | 785. Is Graph Bipartite |
90. Subsets II | 1560A - Dislike of Threes |
36. Valid Sudoku | 557. Reverse Words in a String III |
566. Reshape the Matrix | 167. Two Sum II - Input array is sorted |
387. First Unique Character in a String | 383. Ransom Note |
242. Valid Anagram | 141. Linked List Cycle |
21. Merge Two Sorted Lists | 203. Remove Linked List Elements |
733. Flood Fill | 206. Reverse Linked List |
83. Remove Duplicates from Sorted List | 116. Populating Next Right Pointers in Each Node |
145. Binary Tree Postorder Traversal | 94. Binary Tree Inorder Traversal |
101. Symmetric Tree | 77. Combinations |
46. Permutations | 226. Invert Binary Tree |