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:
n_tests = int(input())
for _ in range(n_tests):
    n_vertices, l, r = list(map(int, input().split()))
    index = 0
    i_v = None
    for i_v in range(1, n_vertices):
        n_indexes_here = (n_vertices - i_v) * 2
        if l <= index + n_indexes_here:
            break
        else:
            index += (n_vertices - i_v) * 2
    else:
        print(1)
        continue
    index += 1
    next_print = i_v + 1
    while True:
        if index > r:
            break
        if index % 2 == 1:
            if index >= l:
                print(str(i_v), end=" ")
        else:
            if index >= l:
                print(next_print, end=" ")
            next_print += 1
            if next_print > n_vertices:
                i_v += 1
                if i_v == n_vertices:
                    i_v = 1
                next_print = i_v + 1
        index += 1
    print("")