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

def minp():
	return sys.stdin.readline().strip()

def mint():
	return int(minp())

def mints():
	return map(int, minp().split())

def stupid(x, n):
	k = (n-1)*2
	r = 0
	while x >= k:
		x -= k
		r += 1
		k -= 2
	return r

def findl(x, n):
	l, r = 0, n
	while r - l > 1:
		c = (l + r) // 2
		if (2*n - 1 - c)*c <= x:
			l = c
		else:
			r = c
	#if l != stupid(x, n):
	#	print(l, stupid(x, n), x, n);
	#	raise 123
	return l

def solve():
	n, l, r = mints()
	res = []
	f = (r == n*(n-1) + 1)
	r -= l + f - 1
	l -= 1
	x = findl(l, n)
	l -= (2*n-1-x)*x
	k = (n-1-x)*2
	x += 1
	while False:#l >= k:
		print(l, k)
		l -= k
		x += 1
		k -= 2
	y = x + 1 + l // 2
	if l % 2:
		while r >= 2:
			res.append(y)
			if y == n:
				x += 1
				y = x + 1
			else:
				y += 1
			res.append(x)
			r -= 2
		if r != 0:
			res.append(y)
	else:
		while r >= 2:
			res.append(x)
			res.append(y)
			if y == n:
				x += 1
				y = x + 1
			else:
				y += 1
			r -= 2
		if r != 0:
			res.append(x)
	if f:
		res.append(1)
	print(' '.join(map(str,res)))

for i in range(mint()):
	solve()