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
input = sys.stdin.buffer.readline
def prog():
    for _ in range(int(input())):
        n,l,r = map(int,input().split())
        l -= 1
        r -= 1
        a = 1
        new = 2*(n-a)
        while new < l:
            l -= new
            r -= new
            a += 1
            new = 2*(n-a)
        total = new - l - (2*(n-a))
        if l == 0 and a == 1:
            segment = [1]
            l += 1
            r += 1
        else:
            segment = []
        while total < r and a <= n:
            segment_a = []
            for i in range(1+a,n):
                segment_a.extend([i,a])
            if n > 1+a:
                segment_a.extend([n,1+a])
            segment.extend(segment_a)
            total += 2*(n-a)
            a += 1
        if a == n+1 or n == 2:
            segment.extend([n,1])
        sys.stdout.write(' '.join(map(str,segment[l-1:r])) + '\n')
prog()