Yeah, we failed to make up a New Year legend for this problem.

A permutation of length $n$ is an array of $n$ integers such that every integer from $1$ to $n$ appears in it exactly once. 

An element $y$ of permutation $p$ is reachable from element $x$ if $x = y$, or $p_x = y$, or $p_{p_x} = y$, and so on. 

The decomposition of a permutation $p$ is defined as follows: firstly, we have a permutation $p$, all elements of which are not marked, and an empty list $l$. Then we do the following: while there is at least one not marked element in $p$, we find the leftmost such element, list all elements that are reachable from it in the order they appear in $p$, mark all of these elements, then cyclically shift the list of those elements so that the maximum appears at the first position, and add this list as an element of $l$. After all elements are marked, $l$ is the result of this decomposition.

For example, if we want to build a decomposition of $p = [5, 4, 2, 3, 1, 7, 8, 6]$, we do the following:  initially $p = [5, 4, 2, 3, 1, 7, 8, 6]$ (bold elements are marked), $l = []$;  the leftmost unmarked element is $5$; $5$ and $1$ are reachable from it, so the list we want to shift is $[5, 1]$; there is no need to shift it, since maximum is already the first element;  $p = [\textbf{5}, 4, 2, 3, \textbf{1}, 7, 8, 6]$, $l = [[5, 1]]$;  the leftmost unmarked element is $4$, the list of reachable elements is $[4, 2, 3]$; the maximum is already the first element, so there's no need to shift it;  $p = [\textbf{5}, \textbf{4}, \textbf{2}, \textbf{3}, \textbf{1}, 7, 8, 6]$, $l = [[5, 1], [4, 2, 3]]$;  the leftmost unmarked element is $7$, the list of reachable elements is $[7, 8, 6]$; we have to shift it, so it becomes $[8, 6, 7]$;  $p = [\textbf{5}, \textbf{4}, \textbf{2}, \textbf{3}, \textbf{1}, \textbf{7}, \textbf{8}, \textbf{6}]$, $l = [[5, 1], [4, 2, 3], [8, 6, 7]]$;  all elements are marked, so $[[5, 1], [4, 2, 3], [8, 6, 7]]$ is the result. 

The New Year transformation of a permutation is defined as follows: we build the decomposition of this permutation; then we sort all lists in decomposition in ascending order of the first elements (we don't swap the elements in these lists, only the lists themselves); then we concatenate the lists into one list which becomes a new permutation. For example, the New Year transformation of $p = [5, 4, 2, 3, 1, 7, 8, 6]$ is built as follows:  the decomposition is $[[5, 1], [4, 2, 3], [8, 6, 7]]$;  after sorting the decomposition, it becomes $[[4, 2, 3], [5, 1], [8, 6, 7]]$;  $[4, 2, 3, 5, 1, 8, 6, 7]$ is the result of the transformation. 

We call a permutation good if the result of its transformation is the same as the permutation itself. For example, $[4, 3, 1, 2, 8, 5, 6, 7]$ is a good permutation; and $[5, 4, 2, 3, 1, 7, 8, 6]$ is bad, since the result of transformation is $[4, 2, 3, 5, 1, 8, 6, 7]$.

Your task is the following: given $n$ and $k$, find the $k$-th (lexicographically) good permutation of length $n$.


-----Input-----

The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Then the test cases follow. Each test case is represented by one line containing two integers $n$ and $k$ ($1 \le n \le 50$, $1 \le k \le 10^{18}$).


-----Output-----

For each test case, print the answer to it as follows: if the number of good permutations of length $n$ is less than $k$, print one integer $-1$; otherwise, print the $k$-th good permutation on $n$ elements (in lexicographical order).


-----Example-----
Input
5
3 3
5 15
4 13
6 8
4 2

Output
2 1 3 
3 1 2 5 4 
-1
1 2 6 3 4 5 
1 2 4 3

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = '3\n\nfrom math import factorial as fact\n\nN = 55\nc = [1]\nfor i in range(N):\n\tc.append(fact(i))\ndp = [0] * N\ndp[0] = 1\nfor i in range(1, N):\n\tfor j in range(i):\n\t\tdp[i] += dp[j] * c[i - j - 1]\n\ndef get_kth_cycle(n, k):\n\tif n == 1:\n\t\treturn [1]\n\tans = [-1] * n\n\tans[0] = n - 1\n\tfin = [i for i in range(n)]\n\tfin[0] = n - 1\n\tinit = [i for i in range(n)]\n\tinit[n - 1] = 0\n\tused = [False] * n\n\tused[n - 1] = True\n\tfor i in range(1, n - 1):\n\t\tj = 0\n\t\tcur = fact(n - i - 2)\n\t\twhile True:\n\t\t\twhile used[j] or (i < n - 1 and j == init[i]):\n\t\t\t\tj += 1\n\t\t\tif k > cur:\n\t\t\t\tk -= cur\n\t\t\t\tj += 1\n\t\t\telse:\n\t\t\t\tfin[init[i]] = fin[j]\n\t\t\t\tinit[fin[j]] = init[i]\n\t\t\t\tans[i] = j\n\t\t\t\tused[j] = True\n\t\t\t\tbreak\n\tans[-1] = init[-1]\n\treturn [x + 1 for x in ans]\n\ndef f(n, k):\n\tif n == 0:\n\t\tassert k == 1\n\t\treturn []\n\tcl = 1\n\twhile c[cl - 1] * dp[n - cl] < k:\n\t\tk -= c[cl - 1] * dp[n - cl]\n\t\tcl += 1\n\trest = f(n - cl, (k - 1) % dp[n - cl] + 1)\n\trest = [x + cl for x in rest]\n\tk = (k - 1) // dp[n - cl] + 1\n\treturn get_kth_cycle(cl, k) + rest\n\ndef solve():\n\tn, k = list(map(int, input().split()))\n\tif k > dp[n]:\n\t\tprint(-1)\n\t\treturn\n\tprint(*f(n, k))\n\ndef main():\n\tt = int(input())\n\twhile t > 0:\n\t\tt -= 1\n\t\tsolve()\n\nmain()\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 19315