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
input = sys.stdin.readline
from collections import *

def binary_search1():
    l, r = 1, n-1
    
    while l<=r:
        m = (l+r)//2
        
        if acc[m]-acc[1]+1<=left:
            l = m+1
        else:
            r = m-1
    
    return r
    
def binary_search2():
    l, r = 1, n-1
    
    while l<=r:
        m = (l+r)//2
        
        if acc[m+1]-acc[1]>=right:
            r = m-1
        else:
            l = m+1
    
    return l

T = int(input())

for _ in range(T):
    n, left, right = map(int, input().split())
    l = [0, 2*n-2]
    
    for _ in range(n-2):
        l.append(l[-1]-2)
    
    l[-1] += 1
    #print(l)
    acc = [0]
    
    for li in l:
        acc.append(acc[-1]+li)
    
    left_n = binary_search1()
    right_n = binary_search2()
    #print(left_n)
    #print(right_n)
    ans = []
    
    for i in range(left_n, right_n+1):
        li = []
        
        for j in range(l[i]-1):
            if j%2==0:
                li.append(i)
            else:
                if j==1:
                    li.append(i+1)
                else:
                    li.append(li[-2]+1)
        
        if i==n-1:
            li.append(1)
        else:
            li.append(n)
            
        ans += li
    #print(ans)
    sta = left-(acc[left_n]-acc[1])-1
    #print(sta)
    print(*ans[sta:sta+right-left+1])