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

from bisect import bisect_left

T = int(input())
for _ in range(T):
    n, l, r = map(int, input().split())
    ls = [ (n-u)*2 for u in range(n+1) ]
    ls[0] = 1
    for i in range(1, n+1):
        ls[i] += ls[i-1]

    p = bisect_left(ls, l)
    sp = []
    if p == 0:
        sp = [0, 0, 0]
    else:
        d = l - ls[p-1]
        sp = [p, p+(d+1)//2, 0 if d%2 else 1]
    
    def getp(ls):
        if ls[0] == 0 or ls == [n-1, n, 1]: return 1
        elif ls[2] == 0: return ls[1]
        elif ls[1] == n: return ls[0]+1
        else: return ls[0]

    def nextp(ls):
        a, b, c = ls
        if a == 0: return [1, 2, 0]
        elif c == 0: return [a, b, 1]
        elif b < n: return [a, b+1, 0]
        else: return [a+1, a+2, 0]
         
    res = []
    for _ in range(r-l+1):
        #print(sp)
        res.append(getp(sp))
        sp = nextp(sp)
    print(' '.join(map(str, res)))