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 os, sys, bisect, copy
from collections import defaultdict, Counter, deque
from functools import lru_cache   #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------

for _ in range(int(input())):
    n,l,r = mapi()
    low = 1
    high = n-1
    pst = 0
    def check(pst,n,l):
        val = 2*n*pst-(pst*pst)-pst
        if val < l:
            return True
        return False
    while low <= high:
        mid = (low+high)//2
        if check(mid,n,l):
            pst = mid
            low = mid+1
        else:
            high = mid-1
    val = 2*n*pst-(pst*pst)-pst
    rem = l-val
    #print(pst)
    temp = True
    prev = True
    for i in range(rem-1):
        if prev:
            if temp:
                pst += 1
                nxt = pst
        else:
            nxt += 1
            temp = False
        if nxt == n:
            temp = True
        prev = not prev
        #print(pst,nxt)
    #print(temp,prev,nxt)
    res = []
    for i in range(r-l+1):
        if prev:
            if temp:
                pst += 1
                nxt = pst
            if pst == n:
                pst = 1
            res.append(pst)
        else:
            temp = False
            nxt += 1
            res.append(nxt)
        if nxt == n:
            temp = True
        prev = not prev
    print(*res)