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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.ArrayList;
import java.util.List;

public class CE85D {
	public static void main(String[] args) throws NumberFormatException, IOException {
		FastReader sc=new FastReader();
	
		int t = sc.nextInt();
		for(int x=0; x<t; x++) {
			long n = sc.nextLong();
			long l = sc.nextLong();
			long r = sc.nextLong();
			
			long total = (n)*(n-1)+1;
			
			//System.out.println("totes" + total);
			
//			if(l == total && r==total) {
//				System.out.println(1);
//				continue;
//			}
			
			long first = 1;
			long second = 0;
			
			long sum = 0;
			
			while(true) {
				long next = 2*(n-first);
				//System.out.println(next);
				if(sum + next < l) {
					first += 1;
					sum += next;
				} else {
					break;
				}
				
				if(next == 2) {
					break;
				}
			}
			
			//System.out.println("sumdy" + sum);
			
			long diff = l - sum;
			second = first + 1 + ((diff-1)/2);
			
			//System.out.println("fgege" + first + " " + second);
			
			while(l <= r) {
				//System.out.println("ell"+ l);
				if(l == total) {
					System.out.print("1 ");
				} else {
					if(l%2 == 0) { //second
						System.out.print(second + " ");
						if(second < n) {
							second += 1;
						} else {
							first += 1;
							second = first + 1;
						}
					} else {
						System.out.print(first + " ");
					}
				}
				l++;
			}
			System.out.println();
		}
	}
	
	static class FastReader{BufferedReader br;StringTokenizer st;public FastReader()
	{br=new BufferedReader(new InputStreamReader(System.in));}
	String next(){while(st==null||!st.hasMoreElements())
	{try{st=new StringTokenizer(br.readLine());}
	catch(IOException e){e.printStackTrace();}}return st.nextToken();}
	int nextInt(){return Integer.parseInt(next());}
	long nextLong(){return Long.parseLong(next());}
	double nextDouble(){return Double.parseDouble(next());}
	String nextLine(){String str="";try{str=br.readLine();}
	catch(IOException e){e.printStackTrace();}return str;}}
}

/* 
int n = sc.nextInt();

for(int y=0; y<n; y++){
	int x = sc.nextInt();
}

int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int d = sc.nextInt();
int e = sc.nextInt();
*/