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:
import sys

#  2 -> 1 2 1
#  3 -> 1 2 1 3 2 3 1
#  4 -> 1 2 1 3 1 4 2 3 2 4 3 4 1
#  5 -> 1 2 1 3 1 4 1 5  2 3 2 4 2 5  3 4 3 5  4 5  1

# for X(k, 1): first 2(k-1) are  1, i; then (k-1)(k-2) are X(k-1, 2) ; then 1

# 2(k-1), 2(k-2), 2(k-3) ...


def binsearch(i, n):

    last = n*(n-1)
    ip = n*(n-1) - i
    
    #print(f"binsearch({i},{n}) last={last} ip={ip}")
    jprev = 0
    j = n-1

    while True:
        #print(f"j={j}, jprev={jprev}")
        #print(f"{j} {last-j*(j-1)} {last-(j-2)*(j-1)} ")
        step = max(1, abs(jprev-j)//2)
        jprev = j
        if j*(j-1) < ip:
            j += step
        elif j > 1 and (j-1)*(j-2) >= ip:
            j -= step
        else:
            break

    k = i - (last - j*(j-1))
    return j, k


def segment(n, l, r):

    res = []
    j1, k1 = binsearch(l-1, n)
    j2, k2 = binsearch(r-1, n)

    # print(f"from {j1, k1} to {j2, k2}")

    for j in range(j1, j2-1, -1):
        for k in range(2*(j-1)):
            if not (j==j1 and k<k1) and not (j==j2 and k>k2):
                if k % 2 == 0:
                    res.append(n-j+1)
                else:
                    res.append(k // 2 + n-j + 2)
    if j2 == 1 and k2 == 0:
        res.append(1)
    return res

t = int(input().strip())

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

    print(' '.join(map(str, segment(n, l, r))))