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 io, os
def main():
    input=sys.stdin.readline
    t=int(input())
    for i in range(t):
        n,l,r=map(int,input().split())
        end=[]
        start=(n-1)*2
        copy=0
        count,ok=0,True
        for i in range(1,n+1):
            if (count+start)<l:
                count+=start
                copy+=start
                start-=2
            else:
                pq=count
                for j in range(i+1,n+1):
                    end.append(i)
                    end.append(j)
                    pq+=2
                    count=pq
                    if pq>=r:
                        ok=False
                        break
            if ok==False:
                break
        if r==n*(n-1)+1:
            end.append(1)
        #print(end,copy)
        ans=' '.join(map(str,end[l-copy-1:r-copy]))
        sys.stdout.write(ans+'\n')
        
        
                  

            
if __name__ == "__main__": 
    main()