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.BufferedInputStream;
import java.util.Scanner;

public class Main {

    private static Scanner sc = new Scanner(new BufferedInputStream(System.in));

    public static void main(String[] args) {
        work();
    }

    private static void work() {
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            long n = sc.nextLong();
            long l = sc.nextLong(), r = sc.nextLong();
            printResult(n, l, r);
        }
    }

    private static void printResult(long n, long l, long r) {
        StringBuffer stringBuffer = new StringBuffer();
        long k = (long) ((double) (2 * n - 1) - Math.sqrt((2 * n - 1) * (2 * n - 1) - 4 * l)) / 2;
        int cnt = 0;
        long  p = l, bias;
        while (cnt < r - l + 1) {
            if (p == n * (n - 1) + 1) {
                stringBuffer.append(1).append(' ');
                break;
            }
            bias = p - (2 * n - k - 1) * k;
            if (bias != 0) {
                if (bias % 2 == 1)
                    stringBuffer.append(k + 1).append(' ');
                else
                    stringBuffer.append(bias / 2 + k + 1).append(' ');
            } else {
                k--;
                stringBuffer.append(n).append(' ');
            }
            p++;
            cnt++;
            if (p > (2 * n - k - 2) * (k + 1))
                k++;
        }
        System.out.println(stringBuffer);
    }

}