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 A
{
      public static void main(String ar[]) throws Exception
      {
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            int t=Integer.parseInt(br.readLine());
            for(int p=0;p<t;p++)
            {
                  String s1[]=br.readLine().split(" ");
                  int n=Integer.parseInt(s1[0]);
                  long l=Long.parseLong(s1[1]);
                  long r=Long.parseLong(s1[2]);
                  
                  
                  int u=1;
                  long x=0;
                  while(x<l && u<n)
                  {
                      x+=2*n-2*u;
                      if(x<l)
                       u++;
                      else
                      { x-=2*n-2*u; break; }
                  }
                  
                 //System.out.println(x+" "+u);
                  StringBuffer sb=new StringBuffer();
                  while(x<r)
                  {
                        for(int i=u+1;i<=n;i++)
                        {
                              x++;
                              if(x>=l && x<=r)
                               sb.append(u).append(" ");
                              x++;
                              if(x>=l && x<=r)
                               sb.append(i).append(" ");
                              // if(i!=n || (i==n && u==n-1))
                              //  x++;
                              // if(i!=n && x>=l && x<=r)
                              // System.out.println(u);
                              // if(i==n && u==n-1 && x+1>=l && x+1<=r)
                              // { System.out.print("1 "); x++; }
                        }
                        u++;
                        if(u>=n && x+1>=l && x+1<=r)
                        { sb.append("1 "); x++; }
                        //System.out.println("fcgjh");
                  }
                  
                  //System.out.println(sb);
                  System.out.println(sb);
            }
      }
}