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
input = sys.stdin.buffer.readline

for t in range(int(input())):
    n,l,r = map(int,input().split())
    for i in range(l,min(2*(n-2)+1,r) + 1):
        print('1' if i & 1 else i//2 + 1 , end = ' ')
    n_set = n
    set_idx = 2*(n-2) + 2
    
    while(n_set > 2):
        ls = l - set_idx + 1
        rs = r - set_idx + 1
        set_idx += 2*(n_set - 2)
        set_par = n - n_set + 2
        if ls < 2 and rs >= 1:print(n , end = ' ')
        for i in range(max(2,ls) , min(2*(n_set - 2),rs) + 1):
            print(set_par + (i-1)//2 if i & 1 else set_par , end = ' ')
        n_set -= 1
    l -= set_idx
    r -= set_idx
    if l <= 0 and r>=0:print(n , end = ' ')
    if r == 1:         print(1 , end = ' ')
    print()