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 javax.management.MBeanRegistration;
import java.io.*;
import java.util.*;

public class Main {
    private void solve() {
        long n = sc.nextLong();
        long l = sc.nextLong();
        long r = sc.nextLong();

        long last = n * (n - 1) + 1;
        if (l == last) {
            out.println(1);
            return;
        }

        long g = 1;
        long k = 0;
        while (g < n) {
            long m = (n - g) * 2;
            if (k + m >= l) {
                break;
            }
            k += m;
            ++g;
        }

        long start = g + 1;
        for (long i = k + 1; i <= r; i += 2) {
            if (i == last) {
                out.println(1);
                break;
            }

            if (i >= l) {
                out.print(g);
                out.print(i == r ? "\n" : " ");
            }

            if (i + 1 >= l && i + 1 <= r) {
                out.print(start);
                out.print(i + 1 == r ? "\n" : " ");
            }

            if (start == n) {
                ++g;
                start = g + 1;
            } else {
                ++start;
            }
        }
    }

    private void run() {
        int T = sc.nextInt();
        for (int t = 0; t < T; ++t) {
            solve();
        }
    }

    private static PrintWriter out;
    private static MyScanner sc;

    private static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        private MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

    public static void main(String[] args) {
        out = new PrintWriter(new BufferedOutputStream(System.out));
        sc = new MyScanner();
        new Main().run();
        out.close();
    }
}