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_LINES = sys.stdin.read().splitlines()
# input = iter(_INPUT_LINES).__next__





def go():
    # n=int(input())
    n,l,r = map(int, input().split())
    # a = sorted(map(int, input().split()),reverse=True)
    tot = n*(n-1)+1
    add=[]
    if r==tot:
        add=['1']
        r-=1

    res=[]
    if l<=r:

        cur=0
        nextg=n-1
        while cur+2*nextg<l:
            cur+=2*nextg
            nextg-=1

        g=n-nextg
        shift=l-cur-1
        pair,par = divmod(shift,2)
        pair+=g+1

        while l<=r:
            if par==0:
                res.append(g)
                par+=1
            else:
                res.append(pair)
                par=0
                if pair<n:
                    pair+=1
                else:
                    g+=1
                    pair=g+1
            l+=1

    res = res+add
    return ' '.join(map(str,res))


# x,s = map(int,input().split())
t = int(input())
# t = 1
ans = []
for _ in range(t):
    # print(go())
    ans.append(str(go()))
#
print('\n'.join(ans))