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 collections import *
import sys
try: inp = raw_input
except: inp = input
def err(s):
    sys.stderr.write('{}\n'.format(s))

def ni():
    return int(inp())

def nl():
    return [int(_) for _ in inp().split()]


def solve():
    n, L, R = nl()
    L -= 1
    R -= 1
    SM = 0
    out = []
    for i in range(1, n):
        no = (n - i)*2
        if no + SM <= L:
            SM += no
            continue
        if SM > R: continue
        ARR = [i if j%2 == 0 else i + j//2 + 1 for j in range(no)]
        start = max(0, L - SM)
        end = R - SM + 1
        out.extend(ARR[start:end])
        SM += no
    if R == SM:
        out.append(1)
    print(' '.join(map(str, out)))





T = ni()
for _ in range(T):
    solve()