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 PYTHON):

Solution:
rr = raw_input
rri = lambda: int(rr())
rrm = lambda: map(int, rr().split())

def solve(N, L, R):
    L -= 1; R -= 1
    if N == 1:
        return [1][L:R+1]
    if N == 2:
        return [1,2,1][L:R+1]
    if N == 3:
        return [1,2,1,3,2,3,1][L:R+1]

    page = 2*N - 2
    k = 1
    left = 0
    while left + page <= L:
        left += page
        k += 1
        page -= 2
        if page == 0: page = 1

    def write(k):
        for j in xrange(k+1, N+1):
            bns.append(k)
            bns.append(j)
        if k == N:
            bns.append(1)

    bns = []
    while len(bns) < R - left + 1:
        write(k)
        k += 1
    return bns[L-left:R-left+1]
    

for tc in xrange(rri()):
    ans = solve(*rrm())
    print " ".join(map(str, ans))