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.*;
import java.util.*;

public class D {
	Reader source;
	BufferedReader br;
	StringTokenizer in;
	PrintWriter out;

	public String nextToken() throws Exception {
		while (in == null || !in.hasMoreTokens()) {
			in = new StringTokenizer(br.readLine());
		}
		return in.nextToken();
	}
	public int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}
	public long nextLong() throws Exception {
		return Long.parseLong(nextToken());
	}
	public double nextDouble() throws Exception {
		return Double.parseDouble(nextToken());
	}
	
	public void solve() throws Exception {
		int t = nextInt();
		for (int i = 0; i < t; i++) {
			int n = nextInt();
			long l = nextLong();
			long r = nextLong();
			int a = 1;
			while (a < n && l > 2 * (n - a)) {
				l = l - 2 * (n - a);
				r = r - 2 * (n - a);
				a++;
			}
			if (a == n) a = 1;
			int b = a + 1;
			while (l > 2) {
				l = l - 2;
				r = r - 2;
				b++;
			}
			//out.println("a=" + a + " b=" + b + " l=" + l + " r=" + r);
			while (l <= r) {
				if (l % 2 == 1) {
					out.print(a + " ");
				} else {
					out.print(b + " ");
					b++;
					if (b > n) {
						a++;
						if (a == n) a = 1;
						b = a + 1;
					}
				}
				l++;
			}
			out.println();
		}
	}

	public void run() throws Exception {
		source = OJ ? new InputStreamReader(System.in) : new FileReader("D.in");
		br = new BufferedReader(source);
		out = new PrintWriter(System.out);
		solve();
		out.flush();
	}
	public static void main(String[] args) throws Exception {
		new D().run();
	}
	private boolean OJ = System.getProperty("ONLINE_JUDGE") != null;
}