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:
# HEY STALKER
for _ in range(int(input())):
    n, l, r = map(int, input().split())
    z = 0
    idx = 0
    sd = (n-1)*2
    lst = (n*(n-1))+1
    while z < l:
        if sd <= 0:
            z += 1
            break
        idx += 1
        z += sd
        sd -= 2
    m = idx-1
    c = (n-1)*2
    sm = 0
    for t in range(m):
        sm += c
        c -= 2
    sm += 1
    nikal = l-sm
    p = []
    if idx == 1:
        for t in range(2, n+1):
            p.append(1)
            p.append(t)
    else:
        ii = idx
        for t in range(ii+1, n+1):
            p.append(ii)
            p.append(t)
    p.reverse()
    for tg in range(nikal):
        p.pop()
    p.reverse()
    idx += 1
    i1 = idx
    while len(p) < (r-l+1):
        if i1 == n:
            p.append(1)
        for t in range(i1+1, n+1):
            p.append(i1)
            p.append(t)
        i1 += 1
    print(*p[:(r-l+1)])