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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import java.util.Map;
import java.util.StringTokenizer;
import java.util.TreeSet;

import javax.swing.*;

import static java.util.Arrays.binarySearch;
import static java.util.Arrays.copyOfRange;

public class Main {

    public static void main(String[] args) throws Exception {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;

        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);

        Task solver = new Task();
        solver.solve(1, in, out);
        out.close();
    }
}

class Task {

    private static int mod = 1000000007;

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int T = in.nextInt();
        for (int tc = 1; tc <= T; tc++) {
            int n = in.nextInt();
            long l = in.nextLong();
            long r = in.nextLong();
            long cur = 1;
            for (int x = 1; x < n; x++) {
                if (cur > r) {
                    break;
                }
                long rM = cur + 2 * (n - x) - 1;
                if (rM < l) {
                    cur = rM + 1;
                    continue;
                }
                int y = x + 1;
                for (long i = cur; i <= Math.min(rM, r); i += 2) {
                    if (i >= l) {
                        out.print(x + " ");
                    }
                    if (i + 1 >= l && i + 1 <= r) {
                        out.print(y + " ");
                    }
                    y++;
                }
                cur = rM + 1;
            }
            if (r == n * (n - 1L) + 1) {
                out.print(1);
            }
            out.println();
        }

    }

    // 2 * n 2 * n - 1
    // 1 2 1 3 1 n 2 3 2 4 2 n 3 4 3 5 3 n 1 n - 1 n

}

class InputReader {

    private final BufferedReader reader;
    private StringTokenizer tokenizer;

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

    public String nextLine() {
        try {
            return reader.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(nextLine());
        }
        return tokenizer.nextToken();
    }

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

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

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

}