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 Sol4{
	public static void main(String[] args) throws IOException{
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0) {
			long n = sc.nextInt();
			long l = Long.parseLong(sc.next());
			long r = Long.parseLong(sc.next());
			long idx = 1;
			long cnt = 2*(n-idx);
			while(cnt+1<l) {
				idx++;
				cnt +=(long)2*(n-idx);
			}
			cnt-=2*(n-idx)-1;
			long ix = idx+1;
			while(cnt<r) {
				if(ix == n+1) {
					idx++;
					ix = idx+1;
				}
				if(cnt%2==0) {
					if(cnt>=l)System.out.print(ix + " ");
					ix++;
				}else {
					if(cnt>=l)System.out.print(idx + " ");
				}
				cnt++;
			}
			if(r == ((n)*(n-1)+1))System.out.println(1);
			else {
				if(ix == n+1) {
					idx++;
					ix = idx+1;
				}
				if(cnt%2==0) {
					if(cnt>=l)System.out.println(ix);
				}else {
					if(cnt>=l)System.out.println(idx);
				}
			}
			
			
		}
		sc.close();
	}
}