D. Minimum Euler Cycletime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given a complete directed graph K_n with n vertices: each pair of vertices u \neq v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, \dots, v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 — a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, \dots, v_r.InputThe first line contains the single integer T (1 \le T \le 100) — the number of test cases.Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 \le n \le 10^5, 1 \le l \le r \le n(n - 1) + 1, r - l + 1 \le 10^5) — the number of vertices in K_n, and segment of the cycle to print.It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.OutputFor each test case print the segment v_l, v_{l + 1}, \dots, v_r of the lexicographically smallest cycle that visits every edge exactly once.ExampleInput
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1 
1 3 2 3 
1 
NoteIn the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
3
2 1 3
3 3 6
99995 9998900031 9998900031

1 2 1  1 3 2 3  1  
['constructive algorithms', 'graphs', 'greedy', 'implementation', '*1800']