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.*;
import java.util.*;

public class Main {

    static Parser parser = new Parser();

    public static void main(String[] args) throws IOException {
        int T = parser.parseInt();

        for(int i = 0; i < T; i++){
            solve();
        }
    }

    static void solve() throws IOException{
        int n = parser.parseInt();
        long l = parser.parseLong();
        long r = parser.parseLong();

        int idx = 1;
        long curr = 0;
        while(curr + (n - idx) * 2 < l){
            if(idx == n){
                break;
            }

            curr += (n - idx) * 2;
            idx += 1;
        }
        List<Integer> cycle = new ArrayList<>();
        while(cycle.size() < r - curr + 1){
            if(idx == n){
                break;
            }

            for(int i = idx + 1; i <= n; i++){
                cycle.add(idx);
                cycle.add(i);
            }
            idx += 1;
        }
        cycle.add(1);

        StringBuilder sb = new StringBuilder();
        for(long i = l; i <= r; i++){
            sb.append(cycle.get((int)(i - curr - 1)));
            sb.append(' ');
        }
        System.out.println(sb.toString());

    }

}

class Parser {
    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    private static final Iterator<String> stringIterator = br.lines().iterator();
    private static final Deque<String> inputs = new ArrayDeque<>();

    void fill() throws IOException {
        if(inputs.isEmpty()){
            if(!stringIterator.hasNext()) throw new IOException();
            inputs.addAll(Arrays.asList(stringIterator.next().split(" ")));
        }
    }

    Integer parseInt() throws IOException {
        fill();
        if(!inputs.isEmpty()) {
            return Integer.parseInt(inputs.pollFirst());
        }
        throw new IOException();
    }

    Long parseLong() throws IOException {
        fill();
        if(!inputs.isEmpty()) {
            return Long.parseLong(inputs.pollFirst());
        }
        throw new IOException();
    }

    Double parseDouble() throws IOException {
        fill();
        if(!inputs.isEmpty()) {
            return Double.parseDouble(inputs.pollFirst());
        }
        throw new IOException();
    }

    String parseString() throws IOException {
        fill();
        return inputs.pollFirst();
    }

}