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:
for f in range(int(input())):
    n,l,r=map(int,input().split())
    s=[0]*(r-l+1)
    i=1
    p=1
    t=2*n-((2*n)**2-4*l)**0.5
    t=t/2
    t=int(t)
    t-=1
    if t>0:
        p=2*(t*n-(t*(t+1))//2)
        p+=1
        i+=t
    while p+2*(n-i)<=l and i<n:
        p+=2*(n-i)
        i+=1
    j=i+1
    while p+2<=l:
        p+=2
        j+=1
    sm=1
    if p<l:
        sm=0
    for foo in range(r-l+1):
        if sm==1:
            s[foo]=i
            sm=0
        else:
            s[foo]=j
            j+=1
            sm=1
            if j==n+1:
                i+=1
                j=i+1
    if r==n*(n-1)+1:
        s[r-l]=1
    print(*s)