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.lang.*;
import java.io.*;
public class GFG{     
    
    public static void main (String[] args)  throws Exception
    {
        FastScanner sc = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int t= sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            long l,r;
            l=sc.nextLong();
            r=sc.nextLong();
            l--;
            long s=0;

            for(int v=1;l<r && v<n;v++){
                long len=n-v;
                if(s+2*len<=l){
                    s+=2*len;
                    continue;
                }
                for(int u=v+1;u<=n;u++){
                    if(s==l  && l<r){
                        out.print(v+" ");
                        l++;
                    }
                    s++;
                    if(s==l && l<r){
                        out.print(u+" ");
                        l++;
                    }
                    s++;


                }
            }

            if(l<r){
                out.print("1");
            }   
            out.println();


        }
        
        out.close();

    }
 
}
class FastScanner
{

    final private int BUFFER_SIZE = 1 << 17;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public FastScanner(InputStream in)
    {
        din = new DataInputStream(in);
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }
    public String nextLine() throws Exception
    {
        StringBuffer sb = new StringBuffer("");
        byte c = read();
        while (c <= ' ') c = read();
        do
        {
            sb.append((char)c);
            c = read();
        }
        while(c > ' ');
        return sb.toString();
    }

    public char nextChar() throws Exception
    {
        byte c = read();
        while(c <= ' ') c = read();
        return (char)c;
    }

    public int nextInt() throws Exception
    {
        int ret = 0;
        byte c = read();
        while (c <= ' ') c = read();
        boolean neg = c == '-';
        if (neg) c = read();
        do
        {
            ret = ret * 10 + c - '0';
            c = read();
        }
        while (c > ' ');
        if (neg) return -ret;
        return ret;
    }

    public long nextLong() throws Exception
    {
        long ret = 0;
        byte c = read();
        while (c <= ' ') c = read();
        boolean neg = c == '-';
        if (neg) c = read();
        do
        {
            ret = ret * 10 + c - '0';
            c = read();
        }
        while (c > ' ');
        if (neg) return -ret;
        return ret;
    } 

    private void fillBuffer() throws Exception
    {
        bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
        if (bytesRead == -1) buffer[0] = -1;
    }

    private byte read() throws Exception
    {
        if (bufferPointer == bytesRead) fillBuffer();
        return buffer[bufferPointer++];
    }
    public double nextDouble() throws Exception
    {
        double ret = 0, div = 1;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();

        do
        {
            ret = ret * 10 + c - '0';
        }
        while ((c = read()) >= '0' && c <= '9');

        if (c == '.')
        {
            while ((c = read()) >= '0' && c <= '9')
            {
                ret += (c - '0') / (div *= 10);
            }
        }

        if (neg)
            return -ret;
        return ret;
    }

    
}