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:
import sys

__author__ = 'ratmir'

alphabet = "abcdefghijklmnopqrstuvwxyz"

def solve(n, a, graph):
    return 1

def execute():
    [t] = [int(x) for x in sys.stdin.readline().split()]
    results = []
    for ti in range(0, t):
        #[n] = [int(x1) for x1 in sys.stdin.readline().split()]
        [n, l, r] = [int(x1) for x1 in sys.stdin.readline().split()]

        # ec = []
        # for i in range(1, n+1):
        #     for j in range(i+1, n+1):
        #         ec.append(i)
        #         ec.append(j)
        # ec.append(1)

        cs = 1
        cis = 0
        while l>cis + 2 * (n-cs) and cs<n:
            #print l, cis
            cis += 2*(n-cs)
            cs += 1
        xx = l - cis - 1
        arr = []
        while r > cis and cs<n:
            for i in range(cs+1, n+1):
                arr.append(cs)
                arr.append(i)
            cis += 2*(n-cs)
            cs += 1
        arr.append(1)
        #print xx
        print(''.join('{} '.format(k) for k in arr[xx:xx+r-l+1]))
    print(''.join('{}\n'.format(k) for k in results))


execute()


# 1 2 1
# 1 2 1 3 2 3 1
# 1 2 1 3 1 4 2 3 2 4 3 4 1
# 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5 1