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:
from sys import stdin, gettrace


def input():
    return stdin.buffer.readline()


t = int(input())
for _ in range(t):
    n, l, r = list(map(int, input().split()))
    count = 0
    i = 1
    while count < l:
        if i == n:
            i = 1
        count += (n - i) * 2
        i += 1
    i -= 1
    count -= (n - i) * 2
    j = i + (l - count + 1) // 2
    length = r - l + 1
    if (l % 2) == 0:
        print(j, end=' ')
        j += 1
        length -= 1
        if j > n:
            i += 1
            j = i + 1
    # print("before while i: {}, j: {}, length is {}".format(i, j, length))
    while length > 0:
        if i == n:
            i = 1
        print(i, end=' ')
        length -= 1
        if length <= 0:
            break
        print(j, end=' ')
        length -= 1
        j += 1
        if j > n:
            i += 1
            j = i + 1
    print()

# 1 2 1 3 1 4 1 5 ... 1 n 2 3 2 4 2 5 ... 2 n 3 1 ... 3 n ... n-1 1 ... n-1 n 1
# 1 2 1 3 1 4 1 5. 2 3 2 4 2 5. 3 4 3 5. 4 5. 1