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:
/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class TaskD {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();
    
    int nTests = Integer.parseInt(br.readLine());
    for (int test=0; test<nTests; test++) {
      String[] s = br.readLine().split(" ");
      long n = Integer.parseInt(s[0]);
      long l = Long.parseLong(s[1]);
      long r = Long.parseLong(s[2]);
      
      if (l==r && l==n*(n-1)+1) {
        sb.append("1\n");
        continue;
      }
      int section = 1;
      long end = 2*(n-section);
      while (l>end) {
        section++;
        end += 2*(n-section);
      }
      long[] ends = new long[(int)n];
      ends[0] = 0;
      for (int i=1; i<n; i++) {
        ends[i] = ends[i-1] + 2*(n-i);
      }
      long upper = r;
      if (r==n*(n-1)+1) {
        upper = r-1;
      }
      for (long i=l; i<=upper; i++) {
        if (i!=l)
          sb.append(" ");
        if (i%2==1) {
          sb.append(section);
        } else {
          long offset = i-ends[section-1];
          sb.append(section + (offset/2));
        }
        if (i==ends[section])
          section++;
      }
      if (r==n*(n-1)+1) {
        sb.append(" 1");
      }
      sb.append("\n");
    }
    System.out.print(sb);
  }
}