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.File;
import java.io.IOException;
import java.util.Scanner;

public final class Main
{
	static final long mod = 998244353l;
	static long gain[];

	public static void main(String[] args) throws IOException
	{
		Scanner in = getScan(args);
		int t = in.nextInt();
		while (t-- > 0)
		{
			long n = in.nextInt();
			long l = in.nextLong();
			long r = in.nextLong();
			long m = 2 * (n - 1);
			long i = 1;
			if (l == n * (n - 1) + 1)
			{
				System.out.println(1);
				continue;
			}
			while (l > m)
			{
				l -= m;
				r -= m;
				i++;
				m = 2 * (n - i);
			}
			recursive(n, l, r, i);
			System.out.println();
		}
	}

	public static void recursive(long n, long l, long r, long curr)
	{
		if (curr == n)
		{
			if (r == 1) System.out.print(1);
			return;
		}
		long m = 2 * (n - curr);
		if (l <= m)
		{
			l = Math.max(1, l);
			long max = Math.min(r, m);
			for (long i = l; i <= max; i++)
			{
				if (i % 2 == 1) System.out.print(curr + " ");
				else System.out.print(curr + i / 2 + " ");
			}
		}
		l -= m;
		r -= m;
		if (r <= 0) return;
		recursive(n, l, r, curr + 1);
	}

	public static int log2nlz(int bits)
	{
		if (bits == 0) return 0; // or throw exception
		return 31 - Integer.numberOfLeadingZeros(bits);
	}

	static Scanner getScan(String[] args) throws IOException
	{
		if (args.length == 0)
		{
			return new Scanner(System.in);
		}
		else
		{
			return new Scanner(new File(args[0]));
		}
	}

}

class Node
{
	int val;
	int deg;
	int par;
	int i;
	long nbChild;

	Node(int i)
	{
		this.i = i;
	}

	@Override
	public String toString()
	{
		return i + " " + val;
	}
}