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.LinkedList;
import java.util.List;
import java.util.Scanner;

/**
 * Codeforces Problem 1334D
 */
public class Solution {

	public static void main(String args[]) {
		try (Scanner scanner = new Scanner(System.in);) {
			int t = scanner.nextInt();
			for (int i = 0; i < t; i++) {
				String solution = solveNext(scanner);
				System.out.println(solution);
			}
		}
		System.exit(0);
	}

	public static String solveNext(Scanner scanner) {
		int num = scanner.nextInt();
		long left = scanner.nextLong();
		long right = scanner.nextLong();
		return new Solution(num, left, right).solve();
	}

	int num;
	long left;
	long right;

	public Solution(int num, long left, long right) {
		this.num = num;
		this.left = left;
		this.right = right;
	}

	public String solve() {
		List<Integer> vertices = new LinkedList<>();
		long currEdge = 1;
		for (int i = 1; i < this.num; i++) {
			long nextCutoff = currEdge + 2 * (this.num - i);
			if (this.left > nextCutoff) {
				currEdge = nextCutoff;
				continue;
			}

			if (currEdge > this.right) {
				break;
			}

			for (int j = i + 1; j <= this.num; j++) {
				if (currEdge >= this.left && currEdge <= this.right) {
					vertices.add(i);
				}
				currEdge++;
				if (currEdge >= this.left && currEdge <= this.right) {
					vertices.add(j);
				}
				currEdge++;
			}
		}

		if (currEdge <= this.right) {
			vertices.add(1);
		}

		StringBuilder sb = new StringBuilder();
		for (Integer vertex : vertices) {
			sb.append(vertex).append(" ");
		}

		return sb.toString();
	}

}