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.math.BigInteger;
import java.util.*;
import java.awt.*;
import java.util.List;

public class Main {
    static int mod = (int) 1e9 + 7;
    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        StringBuilder sb = new StringBuilder();
        int t = sc.nextInt();
        while(t-->0) {
            long n = sc.nextLong(), l = sc.nextLong(), r = sc.nextLong();
            long base, nonBase, startIndex;
            base = (long) Math.floor(l / (2 * n));
            while (((base * 2 * n) - (base * (base + 1)) < l) && (base < n)) base++;

            startIndex = ((base - 1) * 2 * n) - (base * (base - 1));
            for (long j = l; j < r + 1; j++) {
                if (j > ((base * 2 * n) - (base * (base + 1)))) {
                    startIndex = j;
                    base += 1;
                }
                nonBase = base + (j - startIndex + 1) / 2;
                if (j == (n * (n - 1)) + 1) {
                    sb.append(1);
                    break;
                }
                if (j % 2 == 1)sb.append(base+" ");
                else sb.append(nonBase+" ");

            }
            sb.append("\n");
        }
        System.out.println(sb);
    }
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader() { 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()); }
    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}