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.Arrays;
import java.util.Scanner;

public class D {

	public static void main(String[] args) {
		Scanner file = new Scanner(System.in);
		int inputs = file.nextInt();
		while(inputs-->0) {
			int n = file.nextInt();
			long l = file.nextLong();
			long r = file.nextLong();
			long[] changes = new long[n+1];
			changes[1] = 2*(n-1);
			for(int i = 2; i <= n; i++) {
				changes[i] = changes[i-1] + Math.max(2*(n-i), 1);
			}
//			System.out.println(Arrays.toString(changes));
			
			int level = 1;
			while(l > changes[level])
				level++;
			long curr = (l-changes[level-1]+1)/2+level;
			StringBuilder ans = new StringBuilder();
			for(long index = l; index <= r; index++) {
				if(index == changes[n])
					ans.append(1);
				else {
					if(index > changes[level]) {
						level++;
						curr = level+1;
					}
					if(index % 2 == 0) {
						ans.append(curr + " ");
						curr++;
					}
					else {
						ans.append(level + " ");
					}
				}
			}
			System.out.println(ans);
		}
		file.close();
	}

}