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.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class D {

    static class Task {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int t = in.nextInt();
            while ((t--) > 0) {
                long n = in.nextLong();
                long l = in.nextLong();
                long r = in.nextLong();

                long sum = 0;
                List<Long> list = new ArrayList<>();
                for (long i = n - 1; i >= 1; i--) {
                    sum += i * 2;
                    if (sum >= l) {
                        long diff = (sum - l) / 2;
                        long first = n - i; // pair的第一个
                        long second = n - diff;
                        if ((sum - l) % 2 == 1) {
                            // 从first开始
                            list.add(first);
                            if (r == l) {
                                break;
                            }
                            l++;
                        }

                        list.add(second);
                        second++;
                        l++;
                        if (second > n) {
                            first++;
                            if (first == n) {
                                first = 1;
                            }
                            second = first + 1;
                        }

                        for (long j = l; j <= r; j+=2) {
                            list.add(first);
                            if (j == r) {
                                break;
                            }
                            list.add(second);
                            second++;
                            if (second > n) {
                                first++;
                                if (first == n) {
                                    first = 1;
                                }
                                second = first + 1;
                            }
                        }

                        break;
                    }
                }

                if (list.isEmpty()) {
                    list.add(1L);
                }

                for (Long num : list) {
                    System.out.print(num + " ");
                }

                System.out.println();
            }
        }
    }

    private static void solve() {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task task = new Task();
        task.solve(1, in, out);
        out.close();
    }

    public static void main(String[] args) {
        new Thread(null, () -> solve(), "1", 1 << 26).start();
    }

    static class InputReader {

        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

    }

}