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 CodeforcesJava;

import java.io.*;
import java.util.*;

public class Main {

    public void solve(InputProvider in, PrintWriter out) throws IOException {
        int testCount = in.nextInt();
        for (int test = 0; test < testCount; test++) {
            long pointCount = in.nextLong();
            long firstIndex = in.nextLong();
            long lastIndex = in.nextLong();
            long firstGroup = 2 * (pointCount - 1);
            boolean needLast = lastIndex == countElements(firstGroup, pointCount - 1) + 1;
            if (firstIndex < pointCount * (pointCount - 1) + 1) {
                if (needLast) {
                    lastIndex--;
                }
                long minGuess = 0;
                long maxGuess = pointCount - 1;
                while (maxGuess > minGuess + 1) {
                    long middleGuess = (maxGuess + minGuess) / 2;
                    long middleGuessCount = countElements(firstGroup, middleGuess);
                    if (firstIndex <= middleGuessCount) {
                        maxGuess = middleGuess;
                    } else {
                        minGuess = middleGuess;
                    }
                }
                while (countElements(firstGroup, minGuess + 1) < firstIndex) {
                    minGuess++;
                }
                long distance = firstIndex - countElements(firstGroup, minGuess);
                long levelPoint = minGuess + 1;
                long point;
                long pairPoint;
                if ((distance & 1) == 1) {
                    point = levelPoint;
                    pairPoint = (distance + 1) / 2 + levelPoint;
                } else {
                    point = distance / 2 + levelPoint;
                    out.print(point + " ");
                    firstIndex++;
                    if (point == pointCount) {
                        point = levelPoint + 1;
                        pairPoint = point + 1;
                    } else {
                        pairPoint = point + 1;
                        point = levelPoint;
                    }
                }
                boolean printMain = true;
                for (long i = firstIndex; i <= lastIndex; i++) {
                    if (printMain) {
                        out.print(point + " ");
                        printMain = false;
                    } else {
                        out.print(pairPoint + " ");
                        printMain = true;
                        if (pairPoint < pointCount) {
                            pairPoint++;
                        } else {
                            point++;
                            pairPoint = point + 1;
                        }
                    }
                }
            }
            if (needLast) {
                out.print("1 ");
            }
            out.print("\n");
        }
    }

    private long countElements(long first, long count) {
        return (2 * first - 2 * (count - 1)) * count / 2;
    }

    public static void main(String[] args) throws Exception {
        try (InputProvider input = new InputProvider(System.in);
             PrintWriter output = new PrintWriter(System.out)) {
            new Main().solve(input, output);
        }
    }

    public static class InputProvider implements AutoCloseable {

        private final BufferedReader reader;
        private StringTokenizer tokenizer;

        public InputProvider(Reader reader) {
            this.reader = new BufferedReader(reader);
        }

        public InputProvider(InputStream input) {
            reader = new BufferedReader(new InputStreamReader(input));
        }

        public String next() throws IOException {
            if (Objects.isNull(tokenizer) || !tokenizer.hasMoreTokens())
                tokenizer = new StringTokenizer(reader.readLine());
            return tokenizer.nextToken();
        }

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

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

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

        public String nextLine() throws IOException {
            return reader.readLine();
        }

        @Override
        public void close() throws Exception {
            reader.close();
        }

    }

}