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:
#!/usr/bin/env python
# coding: utf-8

# In[48]:


#from __future__ import print_function
#from sys import stdin


# In[52]:


cases = int( input() )


# In[53]:


def ecycle(n,l,r):
    l-=1
    cnt = r - l 
    
    p = n-1
    start = 1
    while(l >= 2*p and p>0):
        l -= 2*p
        p -= 1
        start += 1
        
    if(start==n):
        start = 1
    
    
    flag = (l+1)%2
    nextn = start + 1 + l//2
    
    
    while(cnt>0):
        cnt-=1
        if(flag==1):
            print(start,end=" ")
        if(flag==0):
            print(nextn, end=" ")
            nextn += 1
        if(nextn>n):
            start += 1
            nextn = start + 1
            flag = 0
            
            if(start==n):
                start = 1
        
        flag = 1-flag
        
        
    
    


# In[54]:



while(cases>0):
    n,l,r = map( int, input().split() )
    ecycle(n,l,r)
    cases -= 1


# In[ ]: