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.PrintWriter;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
	static Scanner sc = new Scanner(System.in);
	static PrintWriter writer = new PrintWriter(System.out);

	public static void main(String[] args) {
		int T = sc.nextInt();
		for (int i = 0; i < T; i++) {
			solve();
		}
		writer.flush();
	}

	static void solve() {
		int N = sc.nextInt();
		long L = sc.nextLong() - 1;
		long R = sc.nextLong();
		ArrayList<Integer> ans = new ArrayList<>();
		if (L <= 0) {
			ans.add(1);
		}
		long idx = 1;
		for (int i = 2; i < N; i++) {
			if (L <= idx && idx < R) {
				ans.add(i);
			}
			idx++;
			if (L <= idx && idx < R) {
				ans.add(1);
			}
			idx++;
		}
		if (L <= idx && idx < R) {
			ans.add(N);
		}
		idx++;
		for (int i = 2; i < N && idx < R; i++) {
			long add = (N - i) * 2;
			if (L <= idx + add) {
				if (L <= idx && idx < R) {
					ans.add(i);
				}
				idx++;
				for (int j = i + 1; j < N; j++) {
					if (L <= idx && idx < R) {
						ans.add(j);
					}
					idx++;
					if (L <= idx && idx < R) {
						ans.add(i);
					}
					idx++;
				}
				if (L <= idx && idx < R) {
					ans.add(N);
				}
				idx++;
			} else {
				idx += add;
			}
		}
		if (L <= idx && idx < R) {
			ans.add(1);
		}
		for (int i = 0; i < ans.size(); i++) {
			writer.print(ans.get(i) + (i == ans.size() - 1 ? "\n" : " "));
		}
	}
}