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.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.stream.LongStream;

public class Solution {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        scanner.nextLine();
        for (int i = 0; i < t; i++) {
            solve(scanner, i + 1);
        }
    }

    public static void solve(Scanner in, int id) {
        int n = in.nextInt();
        long l = in.nextLong();
        long r = in.nextLong();

        long visits = 0;

        for(int i = 1; i <= n-1; i++) {
            //Calculate visits
            long visits_i = 2*(n - i);
            if(visits + visits_i < l) {
                visits += visits_i;
                continue;
            }

            print(++visits, l, r, i);
            for(int j = i+1; j <= n-1; j++) {
                print(++visits, l, r, j);
                print(++visits, l, r, i);
            }
            print(++visits, l, r, n);

            if(visits > r) {
                System.out.println();
                return;
            }
        }
        print(++visits, l, r, 1);
        System.out.println();
    }

    private static void print(long next_id, long l, long r, int num) {
        if(next_id >= l && next_id <= r) {
            System.out.print(num + " ");
        }
    }
}