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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class D {
	
	public static void main(String[] args) throws Exception
	{
		new D().run();
	}
	
	LinkedList<Integer> inversionLocations = new LinkedList<Integer>();
	int K;
	int inversions;
	char[] chars;
	ArrayList<Integer> toAdd = new ArrayList<Integer>();
	
	public void run() throws Exception
	{
		BufferedReader file = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(file.readLine());
		int T = Integer.parseInt(st.nextToken());
		PrintWriter pout = new PrintWriter(System.out);
		for(int z = 0;z<T;z++)
		{
			st = new StringTokenizer(file.readLine());
			long K = Long.parseLong(st.nextToken());
			long L = Long.parseLong(st.nextToken());
			long R = Long.parseLong(st.nextToken());
			StringBuilder sb = new StringBuilder("");
			for(long i = L;i<=R;i++)
			{
				sb.append(getIndex(K,i-1));
				sb.append(" ");
			}
			pout.println(sb.toString());
		}
		pout.flush();
	}
	
	public long getIndex(long K, long index)
	{
		if(index == K*(K-1))
			return 1;
		 return getIndexHelper(K, index);
		
	}
	
	public long blockSum(long K, long x)
	{
		if(x == 0)
			return 0;
		long first = (K-1)*2;
		long last = (K-x)*2;
		long terms = ((first - last)/2)+1;
		return (first + last) * terms/2;
	}
	
	public int getBlock(long K, long index)
	{
		long L = 0;
		long R = K+1;
		long M = (R+L)/2;
		long ceil = K+1;
		while(R-L > 1)
		{
			M = ((R+L)/2);
			long sum = blockSum(K, M);
			//System.out.println(L+" "+R+" "+M);
			if(sum >= index)
			{
				ceil = M;
				R = M;
			}else {
				L = M;
			}
		}
		return (int)(ceil);
	}
	
	public long getIndexHelper(long K, long index)
	{
		int block = getBlock(K, index+1)-1;
		//System.out.println(index+"_"+block);
		//System.out.println("K = "+K+", index "+index+" block is "+block);
		long remaining = index - blockSum(K, block);
		//System.out.println("remaining: " + remaining);
		if((remaining & 1) == 0)
			return block+1;
		return (block+1) + (remaining/2 + 1);

	}
	
}