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.*;
import java.io.*;
 
public class Solution{
	static PrintWriter out=new PrintWriter(System.out);
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String[] input=br.readLine().trim().split(" ");
		
		int numTestCases=Integer.parseInt(input[0]);
		while(numTestCases-->0){
		    input=br.readLine().trim().split(" ");
		    int n=Integer.parseInt(input[0]);
    		long l=Long.parseLong(input[1]);
    		long r=Long.parseLong(input[2]);
    		printSequence(n,l,r);
		}
        
		
		out.flush();
		out.close();
	}
	public static void printSequence(int n,long l,long r)
	{
	    ArrayList<Integer> ans=new ArrayList<>();
	    long total=1;
	    for(int i=1;i<n && l<=r;i++){
	        long len=n-i;
	        if(total+2*(len)<=l){
	            total+=2*len;
	            continue;
	        }
	        for(int j=i+1;j<=n;j++){
	            if(total==l && l<=r)
	            {
	                ans.add(i);
	                l++;
	            }
	            total++;
	            if(total==l && l<=r){
	                ans.add(j);
	                l++;
	            }
	            total++;
	        }
	    }
	    if(l<=r){
	        ans.add(1);
	    }
        for(int i=0;i<ans.size();i++){
            out.print(ans.get(i)+" ");
        }
	    out.println();
	}
}