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 = 'MAXN = 55\n\nf = [0] * MAXN\nfac = [0] * MAXN\n\nfac[0] = 1\nfor i in range(1, 51):\n  fac[i] = fac[i - 1] * i\n\nf[0] = 1;\nfor i in range(1, 51):\n  f[i] += f[i - 1]\n  for j in range(2, i + 1):\n    f[i] += fac[j - 2] * f[i - j]\n\ndef my_fac(n):\n  if n <= 0:\n    return 1\n  return fac[n]\n\ndef solve_first(n, k):\n  ret = [0] * (n + 1)\n  ret[1] = n\n  for p in range(2, n + 1):\n    for i in range(1, n + 1):\n      if i in ret or i == p:\n        continue\n      ret[p] = i\n\n      cur = p\n      good = True\n      for fuck in range(0, n - 1):\n        cur = ret[cur]\n        if cur == 0:\n          good = True\n          break\n        if cur == p:\n          good = False\n          break\n      \n      if not good:\n        ret[p] = 0\n        continue\n\n      k1 = my_fac(n - p - 1)\n      if k > k1:\n        k -= k1\n      else:\n        break\n      ret[p] = 0\n  ret.pop(0)\n  assert len(ret) == n\n  return ret;\n\ndef solve(n, k):\n  if k == 1:\n    ret = []\n    for i in range(1, n + 1):\n      ret.append(i)\n    return ret\n\n  tot = 0\n  first = -1\n  for i in range(1, n + 1):\n    if tot + my_fac(i - 2) * f[n - i] >= k:\n      first = i\n      break;\n    tot += my_fac(i - 2) * f[n - i]\n\n  k -= tot\n  cnt1 = my_fac(first - 1)\n  cnt2 = f[n - first]\n  \n  x = k // cnt2 + 1\n  y = k % cnt2\n\n  if y == 0:\n    y = cnt2\n    x -= 1\n\n  ret = solve_first(first, x)\n\n  for v in solve(n - first, y):\n    ret.append(v + first)\n\n  return ret\n\nT = int(input())\n\nfor t in range(0, T):\n  s = input().split()\n  n = int(s[0])\n  k = int(s[1])\n  if (k > f[n]):\n    print("-1")\n  else:\n    ans = solve(n, k)\n    for x in ans:\n      print(x, end=" ")\n    print("")'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 18741