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:
// package EducationalRound85;

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

public class ProblemD {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int test=Integer.parseInt(br.readLine());
        StringBuilder print=new StringBuilder();
        while(test--!=0){
            StringTokenizer st=new StringTokenizer(br.readLine());
            int n=Integer.parseInt(st.nextToken());
            long l=Long.parseLong(st.nextToken());
            long r=Long.parseLong(st.nextToken());
            long sum[]=new long[n+1];
            build(sum);
            int ind=search(sum,l,n);
            if(ind==-1){
                print.append("1\n");
                continue;
            }
            long rem=l-sum[ind-1];
            long end=(1l*n*(n-1))+1;
//            System.out.println(end);
            if(rem%2==0){
                long p=ind;
                long curr=p+rem/2;
                boolean chance=false;
                while(l<=r){
                    if(l==end){
                        print.append("1");
                        break;
                    }
                    if(chance){
                        print.append(p+" ");
                        chance=false;
                    }
                    else{
                        print.append(curr+" ");
                        chance=true;
                        curr++;
                        if(curr>n){
                            p++;
                            curr=p+1;
                        }
                    }
                    l++;
                }
            }
            else{
                long p=ind;
                long curr=p+rem/2+1;
                boolean chance=true;
                while(l<=r){
                    if(l==end){
                        print.append("1");
                        break;
                    }
                    if(chance){
                        print.append(p+" ");
                        chance=false;
                    }
                    else{
                        print.append(curr+" ");
                        chance=true;
                        curr++;
                        if(curr>n){
                            p++;
                            curr=p+1;
                        }
                    }
                    l++;
                }
            }
            print.append("\n");
        }
        System.out.print(print.toString());
    }
    public static void build(long sum[]){
        int n=sum.length-1;
        for(int i=1;i<=n;i++){
            long curr=2l*(n-i);
            sum[i]=sum[i-1]+curr;
        }
        return;
    }
    public static int search(long sum[],long l,int n){
        int ans=-1;
        int low=1,high=n-1;
        while(low<=high){
            int mid=(low+high)/2;
            if(sum[mid]>=l){
                ans=mid;
                high=mid-1;
            }
            else{
                low=mid+1;
            }
        }
        return ans;
    }
}