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
import heapq

def input():
    return sys.stdin.readline().strip()
 
def iinput():
    return int(input())
 
def rinput():
    return map(int, input().split())

def rlinput():
    return list(map(int, input().split()))

def YES(flag):
    if flag:
        return "YES"
    return "NO" 

def main():
    def F():
        return l <= w and w <= r
    #n = int(sys.stdin.readline().strip())
    
    n, l, r = rinput()
    res = []
    w = 1
    if F():
        res.append(1)
    for i in range(2, n):
        w += 1
        if F():
            res.append(i)
        w += 1
        if F():
            res.append(1)
    w += 1
    if F():
        res.append(n)
    for i in range(2, n):
        w += 1
        if F():
            res.append(i)
        for o in range(max(w + 1, l), min(w + (n - i - 1) * 2 + 1, r + 1)):
            if (o - w) % 2 == 0:
                res.append(i)
            else:
                res.append((o - w - 1) // 2 + i + 1)
        w += (n - i - 1) * 2 + 1
        if F():
            res.append(n)    
    w += 1
    if F():
        res.append(1)    
    print(*res)
for i in range(iinput()):
    main()