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 JAVA):

Solution:
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		int t=s.nextInt();
		
		StringBuilder sb=new StringBuilder();
		
		for(int i=0;i<t;i++)
		{
			int n=s.nextInt();
			
			long l=s.nextLong();
			long r=s.nextLong();
			
			int left=0;
			int right=0;
			
			long ll=0;
			long rr=0;
			
			long count=0;
			int block=0;
			
			while(count<l&&block<n)
			{
				block++;
				count=count+2*(n-block);
			}
			
			left=block;
			ll=count-2*(n-block);
			
			count=0;
			block=0;
			
			while(count<r&&block<n)
			{
				block++;
				count=count+2*(n-block);
			}
			
			right=block;
			rr=count;
			
			ArrayList<Integer> list=new ArrayList<>();
			
			for(int j=left;j<=right;j++)
			{
				fill(list,j,n);
			}
			
			long cc=0;
			
			for(long j=ll+1;j<=rr;j++)
			{
				int now=list.get((int)(j-(ll+1)));
				
				if(j>=l&&j<=r)
				{
					cc++;
					sb.append(now+" ");
				}
			}
			
			if(cc==r-l+1)
				sb.append("\n");
			else
			{
				sb.append(1+"\n");
			}
			
		}
		
		System.out.println(sb);
	}
	
	public static void fill(ArrayList<Integer> list,int start,int n)
	{
		int c=start+1;
		
		for(int i=0;i<2*(n-start);i++)
		{
			if(i%2==0)
			{
				list.add(start);
			}
			else
			{
				list.add(c);
				c++;
			}
		}
	}
	
}