Problem: 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:

  1. initially p = [5, 4, 2, 3, 1, 7, 8, 6] (bold elements are marked), l = []; 
  2. 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; 
  3. p = [5, 4, 2, 3, 1, 7, 8, 6], l = [[5, 1]]; 
  4. 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; 
  5. p = [5, 4, 2, 3, 1, 7, 8, 6], l = [[5, 1], [4, 2, 3]]; 
  6. 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]; 
  7. p = [5, 4, 2, 3, 1, 7, 8, 6], l = [[5, 1], [4, 2, 3], [8, 6, 7]]; 
  8. 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:

  1. the decomposition is [[5, 1], [4, 2, 3], [8, 6, 7]]; 
  2. after sorting the decomposition, it becomes [[4, 2, 3], [5, 1], [8, 6, 7]]; 
  3. [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 ≤ t ≤ 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 ≤ n ≤ 50, 1 ≤ k ≤ 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long double pi = acos(-1.0);
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const int alf = 26;
const int dig = 10;
const int two = 2;
const int th = 3;
const long long prost = 239;
const long long btc = 30;
const long double eps = 1e-10;
const long long INF = (long long)(1e18 + 239);
const int BIG = (int)(1e9 + 239);
const int MOD = 998244353;
const long long MOD2 = (long long)MOD * (long long)MOD;
mt19937 rnd(239);
const int M = (int)(1e6 + 239);
const int N = (int)(2e3 + 239);
const int L = 20;
const int T = (1 << 20);
const int B = 500;
const int X = 55;
long long sum(long long a, long long b) {
  if (a + b >= INF) return INF;
  return a + b;
}
long long prod(long long a, long long b) {
  if (a == 0 || b == 0) return 0;
  if (a >= INF || b >= INF) return INF;
  if (a <= (INF / b)) return a * b;
  return INF;
}
long long fact[X], dp[X];
vector<int> get_perm(int n, long long k) {
  if (n == 0) return {};
  vector<int> p = get_perm(n - 1, k % fact[n - 1]);
  vector<int> ans = {(int)(k / fact[n - 1])};
  vector<int> val;
  for (int i = 0; i < n; i++)
    if (i != ans[0]) val.push_back(i);
  for (int i : p) ans.push_back(val[i]);
  return ans;
}
bool cmpsec(pair<int, int> a, pair<int, int> b) {
  return (a.second < b.second);
}
vector<int> gett(int n, long long k) {
  if (n == 0) return {};
  if (n == 1) return {0};
  vector<int> ans(n);
  vector<pair<int, int>> al;
  for (int i = 0; i < n - 1; i++) al.push_back(make_pair(i, i));
  for (int i = 1; i < n - 1; i++) {
    pair<int, int> cur;
    for (int t = 0; t < (int)al.size(); t++)
      if (al[t].first == i) {
        swap(al[t], al.back());
        cur = al.back();
        al.pop_back();
        break;
      }
    sort(al.begin(), al.end(), cmpsec);
    long long ost = fact[(int)al.size() - 1];
    int id = k / ost;
    k %= ost;
    ans[i] = al[id].second;
    al[id].second = cur.second;
  }
  ans[n - 1] = al[0].second;
  ans[0] = n - 1;
  return ans;
}
int n;
long long k;
void solve() {
  cin >> n >> k;
  k--;
  if (k >= dp[n]) {
    cout << "-1\n";
    return;
  }
  int was = 0;
  while (n > 0) {
    for (int i = 0; i < n; i++) {
      long long cur = prod(fact[max(0, i - 1)], dp[n - i - 1]);
      if (cur > k) {
        vector<int> nxt = gett(i + 1, k / dp[n - i - 1]);
        for (int t : nxt) cout << t + was + 1 << " ";
        k %= dp[n - i - 1];
        n -= i + 1;
        was += i + 1;
        break;
      }
      k -= cur;
    }
  }
  cout << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fact[0] = 1;
  for (int i = 1; i < X; i++) fact[i] = prod(fact[i - 1], i);
  dp[0] = 1;
  for (int i = 1; i < X; i++) {
    dp[i] = 0;
    for (int j = 0; j < i; j++)
      dp[i] = sum(dp[i], prod(fact[max(0, j - 1)], dp[i - j - 1]));
  }
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}