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 math
#from functools import lru_cache
#import heapq
#from collections import defaultdict
#from collections import Counter
#from sys import stdout
#from sys import setrecursionlimit
from sys import stdin
input = stdin.readline

for ti in range(int(input().strip())):
    n, l, r = [int(x) for x in input().strip().split()]
    cs, tc, cc = 1, 0, n - 1
    ans = []
    while(tc + 2*cc < l and cs<=n):
        tc += 2*cc
        cs += 1
        cc -= 1
    #print(tc)
    while(tc<=r and cs<=n):
        for cci in range(cs + 1, n+1):
            ne = cs
            tc += 1
            if(l<=tc<=r):
                ans.append(ne)
            ne = cci
            tc += 1
            if(l<=tc<=r):
                ans.append(ne)
            if(tc>r): break
        cs += 1
        cc -= 1
    if(tc<r): ans.append(1)
    print(*ans)