#include <bits/stdc++.h>
using namespace std;
using ll = long long;
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int T;
    cin >> T;
    while (T--) {
        ll n, k;
        cin >> n >> k;
        // total number of such "zig-zag" perms is 2^(n−1)
        ll total = 1LL << (n - 1);
        if (k > total) {
            cout << -1 << "\n";
            continue;
        }
 
        // zero‐index k
        k--;
 
        vector<int> ans(n);
        int L = 0, R = n - 1;
        // build from i=1..n
        for (int i = 1; i <= n; i++) {
            // each choice of placing 'i' either at left or right
            // accounts for exactly 2^(n-i) permutations
            ll block = 1LL << (n - i);
            ll idx   = k / block;  // 0 means "put at left", 1 means "put at right"
 
            if (idx == 0) {
                ans[L++] = i;
            } else {
                ans[R--] = i;
            }
            k %= block;
        }
 
        // output
        for (int i = 0; i < n; i++) {
            cout << ans[i] << (i+1<n ? ' ' : '\n');
        }
    }
    return 0;
}
 

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6
3 2
3 3
4 11
4 6
6 39
7 34