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:
// Don't place your source in a package
import javax.swing.*;
import java.lang.reflect.Array;
import java.text.DecimalFormat;
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
import java.util.stream.Stream;


// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int Int() {
            return Integer.parseInt(next());
        }

        long Long() {
            return Long.parseLong(next());
        }

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

        int T=Int();
        for(int t=0;t<T;t++){
            int n=Int();
            long l=Long();
            long r=Long();

            Solution sol=new Solution(out);
            sol.solution(n,l,r);
        }
        out.close();

    }


    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}






class Solution{
    PrintWriter out;
    public Solution(PrintWriter out){
        this.out=out;
    }





    public void solution(int n,long l,long r){
        long cur=2*(n-1);
        long sum=(2+cur)*(n-1)/2;
        if(l>sum){
            out.println(1);
            return;
        }


        //20 18 16 14 12 10 8 ...
        sum=0;
        int first=1;
        while(sum+cur<l){
            sum+=cur;
            cur-=2;
            first++;
        }



        for(long i=l;i<=r;i++){
            if(i%2==1){
                out.print(first+" ");
            }
            else{
                long dis=i-sum;
                out.print((first+dis/2)+" ");
            }
            if(i==sum+cur){
                sum+=cur;
                cur-=2;
                first++;
                if(first==n)first=1;
            }
        }
        out.println();
    }




     //1 2 1 3 1 4 2 3 2 4 3 4 1
    //1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5 1
    //1 2 1 3 1 4 1 5 1 6 2 3 2 4 2 5 2 6 3 4 3 5 3 6 4 5 4 6 5 6 1
    // 1 2 1 3 2 3 1


}







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