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:
#                                               |
#   _` |  __ \    _` |   __|   _ \   __ \    _` |   _` |
#  (   |  |   |  (   |  (     (   |  |   |  (   |  (   |
# \__,_| _|  _| \__,_| \___| \___/  _|  _| \__,_| \__,_|

import sys
import collections

def read_line():
	return sys.stdin.readline()[:-1]
 
def read_int():
	return int(sys.stdin.readline())
 
def read_int_line():
	return [int(v) for v in sys.stdin.readline().split()]

t = read_int()
for i in range(t):
	n,l,r = read_int_line()
	u = r-l+1
	lt = l
	k = n
	while lt>0 and k>0:
		k-=1
		lt -= 2*(k)
	lt += 2*(k)
	k = n-k
	
	j= max(k,1)
	s = 0
	for i in range(1,j):
		s += 2*(n-i)

	lt = l
	lt -= s
	lt += 1
	i = j + lt//2
	ans = []
	if l%2==0:
		ans.append(i)
		i+=1	
	while j<=n and len(ans)<=u:
		if j!=k:	
			i = j+1
		while i<=n and len(ans)<=u:
			ans.append(j)
			ans.append(i)
			i+=1
		j+=1
	ans.append(1)
	print(*ans[:u])