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:
//make sure to make new file!
import java.io.*;
import java.util.*;

public class D85b{
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      int t = Integer.parseInt(f.readLine());
      
      for(int q = 1; q <= t; q++){
         StringJoiner sj = new StringJoiner(" ");
         StringTokenizer st = new StringTokenizer(f.readLine());
      
         long n = Long.parseLong(st.nextToken());
         long l = Long.parseLong(st.nextToken());
         long r = Long.parseLong(st.nextToken());
         
         if(l == (n*(n-1)) + 1){
            out.println(1);
            continue;
         }
         
         long seg = 1;
         long i = 1;
         long interval = (n-seg)*2;
         
         //find start
         while(i <= l){
            i+=interval;
            seg++;
            interval = (n-seg)*2;
         }
         
         seg--;
         interval = (n-seg)*2;
         i-=interval;
         
         while(l <= r && seg < n){
            
            for(long k = l-i; k < interval && l <= r; k++){
               if(k%2 == 0){
                  sj.add("" + seg);
               } else {
                  sj.add("" + ((k/2) + seg + 1));
               }
               l++;
            }
            
            seg++;
            i+=interval;
            interval = (n-seg)*2;
         }
         
         if(l == r) sj.add("" + 1);
         
         
         out.println(sj.toString());
               
         
      }
      
      
      
      
      out.close();
   }
   
      
}