Problem: You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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, ..., 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}, ..., v_r.

Input

The first line contains the single integer T (1 ≤ T ≤ 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 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 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}, ..., 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.

Solution (in PYTHON3):

Solution:
from sys import stdin
input = stdin.buffer.readline
for _ in range(int(input())):
	n, l, r = map(int, input().split())
	l -= 1
	r -= 1
	flag, tmp, s = 1, 0, 0
	ans = []
	x, y = n, n - 1
	for i in range(1, n):
		s += 2 * (n - i)
		if l < s and flag:
			x, tmp = i, (i > 1) * (s - 2 * (n - i))
			l -= tmp
			flag = 0
		if r < s:
			y = i
			r -= tmp
			break
	if x > y:
		l, r = 0, 0
	for i in range(x, y + 1):
		for j in range(i + 1, n + 1):
			ans.append(i)
			ans.append(j)
	#if r >= len(ans) and y != n - 1:
	#	print(0/0)
	ans.append(1)
	print(*ans[l : r + 1])