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:
def ss(a,d,n):

    return n*(2*a+(n-1)*d)//2


T = int(input())

for loop in range(T):

    n,l,r = map(int,input().split())

    ans = []

    lb = 0
    rb = n-1

    while rb - lb != 1:

        m = (lb+rb) // 2

        if ss(2*n-2,-2,m) >= l:
            rb = m
        else:
            lb = m

    BB = rb
    ind = ss(2*n-2,-2,rb-1) + 1
    inind = BB+1
    state = 0
    
    while ind <= r:

        #print (ind,BB,inind)

        now = 0
        if state == 0:
            now = BB
            state = 1
        else:
            now = inind
            inind += 1
            state = 0

            if inind == n+1:
                BB += 1
                inind = BB+1
        
        if ind == ss(2*n-2,-2,n-1) + 1:
            ans.append(1)
            break
        if ind >= l:
            ans.append(now)
        ind += 1
        
    print (*ans)