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.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class MinimumEulerCycle {

        static long getValue(long n, long input, long i) {
            long head;
            if (input == n * (n - 1) + 1) {
                return 1;
            }
            if (i == 1) {
                head = 0;
            } else
                head = (i - 1) * (4 * n - 2 * i) / 2;
            long offset = (input - head);
            long value = offset % 2 != 0 ? i : i + offset / 2;
            return value;
        }

        static long getBlock(long n, long input) {
            if ((2 * n - 2) >= input) {
                return 1;
            }
            double a = Math.ceil((-Math.sqrt(Math.pow(n, 2) * 4 - 4 * n - 4 * input + 1) + 2 * n - 1) / 2);
//            double b = Math.floor((-Math.sqrt(Math.pow(n, 2) * 4 - 4 * n - 4 * input + 1) + 2 * n + 1) / 2);
            return (long)a;
        }

        static void check(long n, long l, long r) {
            for (long i = l; i <= r; i++) {
                long block = getBlock(n, i);
                long value = getValue(n, i, block);
                System.out.print(value + " ");
            }
        }


        public static void main(String[] args) {
            int n;
            Scanner sc = new Scanner(System.in);
            n = sc.nextInt();
            long k, l, r;
            sc.nextLine();
            for (int i = 0; i < n; i++) {
                k = sc.nextLong();
                l = sc.nextLong();
                r = sc.nextLong();
                check(k, l, r);
                sc.nextLine();
            }
        }


}