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())


def query(i, n, x):
    if (x % 2 == 1):
        return i
    else:
        return (i + x // 2)


for _ in range(t):
    n, l, r = map(int, input().split())

    i = 1
    s = 0

    includeOne = False

    if r == n * (n - 1) + 1:
        includeOne = True
        r -= 1
    if l == n * (n - 1) + 1:
        print(1)
        continue

    while s + 2 * (n - i) < l:
        s += 2 * (n - i)
        i += 1

    newS = s
    allIs = [i]
    while newS + 2 * (n - i) < r:
        newS += 2 * (n - i)
        i += 1
        allIs.append(i)

    allIin = 0

    answer = []

    i = allIs[0]

    for x in range(l, r + 1):
        r = query(i, n, x - s)
        if r == n:
            s += 2 * (n - i)
            i += 1
        answer.append(r)

    if includeOne:
        answer.append(1)

    print(*answer)