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.BufferedInputStream;
import java.util.Arrays;
import java.util.Scanner;

/**
 * Created by Harry on 4/10/20.
 */
public class test {
    public static void main(String[] args){
        Scanner scanner = new Scanner(new BufferedInputStream(System.in));
        int T = scanner.nextInt();
        for(int t=0; t<T; t++){
            long n = scanner.nextLong();
            long l = scanner.nextLong();
            long r = scanner.nextLong();
            long step = (n-1)*2;
            long sum = 0;
            while(l>sum){
                sum += step;
                if(l>sum){
                    if(step>2){
                        step -= 2;
                    }
                    else{
                        step = 1;
                    }
                }
            }
            long index = l-(sum-step);
            long count = 0;
            while(count<=(r-l)){
                long node = n - step/2;
                if(node==n){
                    System.out.print(1+" ");
                }
                else{
                    if(index%2==1){
                        System.out.print(node+" ");
                    }
                    else{
                        System.out.print((index/2+node)+" ");
                    }
                }
                index++;
                if(index>step){
                    if(step>2){
                        step-=2;
                    }
                    else{
                        step=1;
                    }
                    index = 1;
                }
                count++;
            }
            System.out.println();
        }
    }
}