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.*;
import java.io.*;
public class d {
	static int n;
	static long l, r;
	public static void main(String[] args) {
		FS sc = new FS();
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		for(int tt = 1; tt <= t; ++tt) {
			n = sc.nextInt();
			l = sc.nextLong();
			r = sc.nextLong();

			// binary search for index i
			int lo = 1, hi = n;
			while(lo < hi) {
				int g = (lo + hi + 1) / 2;

				if(getNumBehind(g) < l) lo = g;
				else hi = g - 1;
			}

			// calculate start j?
			//			System.out.println(lo);

			long idx = getNumBehind(lo) + 1;
			floop : for(int i = lo; ; ++i) {
				if(i == n) {
					pw.print("1 ");
					break;
				}
				for(int j = i + 1; j <= n; ++j) {
					if(idx++ >= l) pw.print(i + " ");
					if(idx > r) break floop;
					if(idx++ >= l) pw.print(j + " ");
					if(idx > r) break floop;
				}
			}

			pw.println();
		}
		pw.flush();
	}
	static long getNumBehind(int i) {
		long out = sum(n - 1);
		out -= sum(n - i);
		return out * 2;
	}
	static long sum(long in) {
		return in * (in + 1) / 2;
	}
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}