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constructive algorithms

graphs

greedy

implementation

*1800

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D. Minimum Euler Cycle

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard 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, $$$v_l, v_{l + 1}, \dots, v_r$$$.

Input

The 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 $$$10^5$$$.

Output

For 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.

Example

Input

3 2 1 3 3 3 6 99995 9998900031 9998900031

Output

1 2 1 1 3 2 3 1

Note

In 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 $$$1$$$.

Codeforces (c) Copyright 2010-2024 Mike Mirzayanov

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