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.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Euler {
    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(f.readLine());
        PrintWriter out = new PrintWriter(System.out);
        for (int t1 = 0; t1 < t; t1++) {
            StringTokenizer tokenizer = new StringTokenizer(f.readLine());
            int n = Integer.parseInt(tokenizer.nextToken());
            long l = Long.parseLong(tokenizer.nextToken()) - 1;
            long r = Long.parseLong(tokenizer.nextToken()) - 1;
            int level = 1;
            long index = 0;
            boolean hadFirst = false;
            while (level < n) {
                long end = index + (n - level) * 2 - 1;
                if (intersect(l, r, index, end)) {
                    for (int i = 0; i < n - level; i++) {
                        if (l <= index + i * 2 && index + i * 2 <= r) {
                            if (hadFirst) {
                                out.print(" ");
                                out.print(level);
                            } else {
                                out.print(level);
                                hadFirst = true;
                            }
                        }

                        if (l <= index + i * 2 + 1 && index + i * 2 + 1 <= r) {
                            if (hadFirst) {
                                out.print(" ");
                                out.print(level + i + 1);
                            } else {
                                out.print(level + i + 1);
                                hadFirst = true;
                            }
                        }
                    }
                }

                index = end + 1;
                level++;
            }
            if (l <= index && index <= r) {
                if (hadFirst) {
                 out.print(" ");
                 out.print(1);
                } else {
                    out.print(1);
                    hadFirst = true;
                }
            }
            out.println();
        }
        out.close();
    }

    private static boolean intersect(long start1, long end1, long start2, long end2) {
        return Math.min(end1, end2) >= Math.max(start1, start2);
    }


}