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 PYTHON3):

Solution:
t=int(input())
for _ in range(t):
    n,l,r=map(int, input().split())
    cycle_size = 2 * (n - 1)
    cycle_start = 1
    cycle_number = 1
    # while True:
    #     if cycle_start + cycle_size >= l:
    #         break
    #     cycle_start += cycle_size
    #     cycle_number += 1
    #     cycle_size -= 2
    # print [l,r]

    while cycle_start + cycle_size <= l and cycle_number != n:
        cycle_start += cycle_size
        cycle_number += 1
        cycle_size -= 2

    # print('cycle_start, cycle_size', cycle_start, cycle_size)
    ans = []
    while l <= r:
        # print('l, cycle_start, cycle_size', l, cycle_start, cycle_size)
        if (l - cycle_start) & 1:
            # print('cycle_size', cycle_size)
            if not cycle_size:
                ans.append(1)
            else:
                ans.append(cycle_number + (l - cycle_start + 1) // 2)
        else:
            # print('cycle_size', cycle_size)
            if not cycle_size:
                ans.append(1)
            else:
                ans.append(cycle_number)
        # print('ans', ans)
        # if (l - cycle_start) // 2
        l += 1
        if (l - cycle_start) >= cycle_size:
            cycle_start += cycle_size
            cycle_size -= 2
            cycle_number += 1
    print(' '.join(str(x) for x in ans))