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.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int t = Integer.parseInt(in.readLine());
        for (int te = 0; te < t; te++) {
            StringTokenizer st = new StringTokenizer(in.readLine());
            long n = Long.parseLong(st.nextToken());
            long l = Long.parseLong(st.nextToken());
            long r = Long.parseLong(st.nextToken());
            long start = 1;
            StringBuilder sb = new StringBuilder();
            for (int i = 1; i <= n; i++) {
                //System.out.println("START BEFORE " + start);
                if (start > r) {
                    break;
                }
                if (start + (n-i)*2 - 1 < l) {
                    start += (n-i)*2;
                    continue;
                }
                //System.out.println("START AFTER " + start);
                for (int k = i+1; k <= n; k++) {
                    if (start >= l && start <= r) {
                        sb.append(i);
                        sb.append(' ');
                    }
                    start++;
                    if (start >= l && start <= r) {
                        sb.append(k);
                        sb.append(' ');
                    }
                    start++;
                }
            }
            if (start == r) {
                sb.append('1');
                sb.append(' ');
            }
            out.println(sb.toString());
        }
        out.close();
    }

}
/*
1
3 1 7


 */