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

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 $10^5$.

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.