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).

Idx	Name	Rating
A	New Year Garland	900
B	Verse For Santa	1300
C	Stack of Presents	1400
D	Santa's Bot	1700
E	New Year Permutations	2700
F	New Year and Handle Change	2800

Educational Codeforces Round 79 (Rated for Div. 2)