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
d = []
def bins(k, n):
    l = 0 
    r = n-2
    while l != r:
        mid = (l+r+1)//2
        if d[mid] <= k:
            l = mid
        else:
            r = mid-1
    # print("debug:", l)
    ex = k - d[l]
    if ex &1:
        f = ex//2
        return f + l + 2
    else:
        return l+1



for q in range(int(sys.stdin.readline())):
    n, l, r = [int(j) for j in sys.stdin.readline().split()]
    mx = n*(n-1)
    d = [0]*(n-1)
    for i in range(1, n-1):
        d[i] = d[i-1] + 2*(n-i)
    # print(d[-15:])
    ret = []
    # print("degug2:",bins(9998900028, n),bins(9998900029, n), bins(9998900030, n), bins(9998900031, n) )
    for i in range(l-1, r):
        ret.append(str(bins(0 if i == mx else i, n)))
    sys.stdout.write(" ".join(ret) + '\n')