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:
from sys import stdout
def solve():
    n, l, r = map(int, raw_input().split())
    l -= 1
    for i in xrange(n - 1):
        t = (n - i - 1) * 2
        if l < t:
            break
        l -= t
        r -= t
    else:
        stdout.write('1\n')
        return
    ans = [0] * (r - l)
    i += 1
    p = i + 1
    sw = 0
    for j in xrange(r):
        if sw:
            if j >= l:
                ans[j-l] = p
            p += 1
            if p > n:
                i += 1
                p = i + 1
                if i == n:
                    i = 1
        else:
            if j >= l:
                ans[j-l] = i
        sw = 1 - sw
    stdout.write(' '.join(map(str, ans)))
    stdout.write('\n')

T = int(raw_input())
for _ in xrange(T):
    solve()