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.Scanner;

public class mineuler {
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		PrintWriter out=new PrintWriter(System.out);
		int t=scan.nextInt();
		for(int tt=0;tt<t;tt++) {
			int n=scan.nextInt();
			long l=scan.nextLong()-1, r=scan.nextLong()-1;
			int startnum1=1;
			long id1=0L, ct=n-1;
			while(id1+ct*2<l) {
				id1+=ct*2;
				ct--;
				startnum1++;
			}
			int printed=0;
			int cur=startnum1, add=1;
			long dif=l-id1;
			while(printed<r-l+1) {
				if(dif<=0) {
					printed++;
					out.print(cur);
					out.print(" ");
				}
				if(cur==startnum1) {
					cur+=add;
					if(cur==n) {
						add=0;
						startnum1++;
						if(startnum1==n) startnum1=1;
					}
				}
				else {
					cur=startnum1;
					add++;
				}
				dif--;
			}
			out.println();
		}
		out.close();
	}
}