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 math
import sys
# input = sys.stdin.readline

T=int(input())
for _ in range(T):
	n,l,r=map(int,input().split())
	size=r-l+1
	run=0
	ans=[]
	for i in range(1,n):
		run+=2*(n-i)
		# print(run)
		if l<=run:
			prev=run-2*(n-i)
			gone=l-prev-1
			size+=gone
			# print(gone)
			cur=0
			now=i
			nex=i+1
			flag=True
			while cur<size:
				if flag:
					if gone>0:
						gone-=1
					else:
						ans.append(now)
					flag=False
				else:
					if gone>0:
						gone-=1
					else:
						ans.append(nex)

					nex+=1
					if nex>n:
						now+=1
						nex=now+1
					flag=True
				cur+=1
			break

	if len(ans)==0:
		ans.append(1)

	if r==n*(n-1)+1:
		ans[-1]=1

	print(*ans)