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 PYTHON):

Solution:
'''input
3
2 1 3
3 3 6
99995 9998900031 9998900031
'''

import sys
import math
from collections import Counter

debug = 1
readln = sys.stdin.readline
#sys.setrecursionlimit(1000000)

def write(s):
    sys.stdout.write(str(s))

def writeln(s):
    sys.stdout.write(str(s))
    sys.stdout.write('\n')

def readint():
    return int(readln().strip())

def readints():
    return map(int, readln().split())

def readstr():
    return readln().strip()

def readstrs():
    return readln().split()

def dbg(*args):
    if debug: print(' '.join(map(str, args)))

def solve(n,l,r):
    start = 0
    ans = []
    l, r = l-1, r-1
    for i in xrange(1, n):
        if l < start: break
        next_start = start + (n-i) * 2
        if l < next_start:
            for cl in range(l, min(r+1, next_start)):
                if cl % 2 == 0:
                    ans.append(i)
                else:
                    ans.append((cl - start + 1) / 2 + i)
            l = min(r, next_start)
        start = next_start

    if l == start:
        ans.append(1)

    return ans

t = readint()
for _ in xrange(t):
    n, l, r = readints()
    writeln(' '.join(map(str, solve(n,l,r))))