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:
from sys import stdin, stdout
import math

def expandList(i,n):
    if i!=n:
        out = []
        for j in range(n-i):
            out.append(i)
            out.append(i+j+1)
        return out
    else:
        return [1]

 
T = int(stdin.readline().rstrip())
for iTest in range(T):
    n,l,r = list(map(int,stdin.readline().rstrip().split(' ')))
    
    listSizes = [2*i for i in range(n-1,-1,-1)]
    listSizes[n-1] = 1
    
    found = False
    a = 0
    i=1
    while l>a+listSizes[i-1]:
        a+=listSizes[i-1]
        i+=1
            
    listi = expandList(i,n)
    ind = l-a-1
    startList = listi[ind:]
    
    while r>a+listSizes[i-1]:
        a+=listSizes[i-1]
        i+=1
        startList+=expandList(i,n)
            
    endList = expandList(i+1,n)
    startList += endList
    startList = startList[:r-l+1]
        
    print(' '.join([str(x) for x in startList]))