1334_D. Minimum Euler Cycle
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.
```python
# -*- coding:utf-8 -*-

"""

created by shuangquan.huang at 2020/7/1

"""

import collections
import time
import os
import sys
import bisect
import heapq
from typing import List



def solve(n, l, r):
    # 1, 2, 1, 3, ..., 1, n
    # 2, 3, 2, 4, ..., 2, n
    # ...
    # n-1, n
    # 1
    
    lo, hi = 1, n
    while lo <= hi:
        k = (lo + hi) // 2
        s = k * (2*n-1-k)
        if s < l:
            lo = k + 1
        else:
            hi = k - 1
    
    k = lo
    s = k * (2*n-1-k)
    b = k
    
    
    # [b, b+1, b, b+2, ..., b, n]
    row = []
    for i in range(b+1, n+1):
        row.append(b)
        row.append(i)
    ans = row[l-s-1:]
    d = r-l+1
    if len(ans) >= d:
        return ans[:d]
    
    while len(ans) < d:
        b += 1
        row = []
        for i in range(b + 1, n + 1):
            row.append(b)
            row.append(i)
        if not row:
            break
        ans += row
    
    ans.append(1)
    # print(ans[:d])
    return ans[:d]
    

if __name__ == '__main__':
    T = int(input())
    ans = []
    for ti in range(T):
        N, L, R = map(int, input().split())
        ans.append(solve(N, L, R))
    print('\n'.join([' '.join(map(str, v)) for v in ans]))
```