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.readline
q = int(input())
for _ in range(q):
	n,l,r = map(int,input().split())
	tab = [-1,1]
	cyk = 0
	for i in range(n):
		cyk += 2*(n-i-1)
		tab.append(cyk + 1)
	tab.append(1012809128301279797787789798789798798)
	i = 0
	start = i
	while True:
		if tab[i + 1] > l:
			start = i
			break
		else:
			i += 1
	#zaczynamy od i X i (x+1) ...
	ind = l
	dlug = 0
	koniec = 0
	while True:
		if koniec:
			break
		target = tab[start+1]
		while ind < target:
			if koniec:
				break
			if ind == (n*(n-1) + 1):
				print(1, end = " ")
				dlug += 1
			else:
				dlug += 1
				if ind%2==1:
					print(i, end = " ")
				else:
					print(n-((target-ind)-1)//2, end = " ")
				ind += 1
			if dlug >= (r-l+1):
				koniec = 1
		start += 1
		i += 1
	print()