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.Scanner;

public class ProblemD {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		
		for(int a=0;a<t;a++) {
			int n = s.nextInt();
			long l = s.nextLong();
			long r = s.nextLong();
			
			long[] arr = new long[n];
			
			long sum = 0;
			long val = 2*(n-1);
			for(int i=0;i<n-1;i++) {
				sum += val;
				val -= 2;
				
				arr[i] = sum;
			}
			arr[n-1] = arr[n-2] + 1;
			
//			for(int i=0;i<n;i++)
//				System.out.println(arr[i]);
			int index = upperBound(arr, l);
			//System.out.println(index);
			print(arr, l, r, index);
			System.out.println();
		}
	}
	
	public static void print(long[] arr, long l, long r, int index) {
		int n = arr.length;
		
		if(index == n-1) {
			System.out.print(1+" ");
			return;
		}
		long val1 = index + 1, val2 = 0;
		long end = arr[index];
		if(l%2 == 0) {
			val2 = n - (end-l)/2;
		}
		else {
			val2 = n - (end-l-1)/2;
		}
		
		for(long i=l;l <= Math.min(end, r);l++) {
			if(l%2 == 1)
				System.out.print(val1+" ");
			else {
				System.out.print(val2+" ");
				val2++;
			}
				
		}
		
		if(end < r)
			print(arr, end + 1, r, index + 1);
	}
	
	public static int upperBound(long[] arr, long v) {
		if(v > arr[arr.length-1])
			return -1;
		if(arr[0] > v)
			return 0;
		int low = 0, high = arr.length - 1;
		
		while(low < high) {
			
			if(low == high - 1) {
				if(arr[low] >= v)
					high = low;
				else
					low = high;
				break;
			}
			int mid = (low + high)/2;
			
			if(arr[mid] >= v)
				high = mid;
			else
				low = mid + 1;
		}
		
		return low;
	}

}