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 PYTHON):

Solution:
for _ in range(input()):
    n,l,r = map(int,raw_input().split())
    i = 1
    s = 0
    while s+(n-i)*2<l:
        s+=(n-i)*2
        i+=1
        if n==i and s<l:
            s+=1
            break
    
    c = 0
    j = i+1
    while c<(r-l+1):
        if c==0:
            for x in range(1,2*(n-i)+1):
                if x+s<l:
                    if x%2==0:
                        j+=1    
                    continue
                
                if c<(r-l+1):
                    if x%2:
                        print i,
                    else:
                        print j,
                        j+=1
                    c+=1
                    
                
        else:
            for x in range(1,2*(n-i)+1):
                if c<(r-l+1):
                    if x%2:
                        print i,
                    else:
                        print j,
                        j+=1
                    c+=1
        if c==r-l and r==n*(n-1)+1:
            print 1,
            break
        i+=1
        j = i+1
    print