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.*;
import java.io.*;

public class Solution {

    public static void main(String[] args) throws IOException{
        
        Scanner in = new Scanner(System.in);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuffer out = new StringBuffer();

        int t=in.nextInt();

        label:
        while(t--!=0) {

            long n=in.nextLong();

            long l=in.nextLong(),
                 r=in.nextLong();

            long cnt=0;
            long i;
            for(i=1L; i<=n; i++) {
                if(l<=cnt+((n-i)<<1))
                    break;
                cnt+=((n-i)<<1);
            }

            for(; i<=n && cnt<=r; i++) {
                for(long j=i+1; j<=n && cnt<=r; j++) {
                    cnt+=1;
                    if(l<=cnt && cnt<=r)
                        out.append(i+" ");
                    cnt+=1;
                    if(l<=cnt && cnt<=r)
                        out.append(j+" ");
                }
            }

            cnt+=1;
            if(l<=cnt && cnt<=r)
                out.append(1+" ");

            out.append("\n");
        }

        System.out.println(out);
    }
}