#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

#define ll long long

const int N = 100005;

ll l[N], r[N];

int find_best(int l1, int r1, int l2, int r2, int need){
    /*
    while(r1 - l1 > 3){
        int m1 = l1 + (r1 - l1) / 3;
        int m2 = r1 - (r1 - l1) / 3;
        int sec = need - m1, diff1, diff2;
        if(l2 > sec)
            diff1 = l2 - sec;
        else if(r2 < sec)
            diff1 = sec - r2;
        else 
            return m1;
        sec = need - m2;
        if(l2 > sec)
            diff2 = l2 - sec;
        else if(r2 < sec)
            diff2 = sec - r2;
        else 
            return m2;
        if(diff1 < diff2)
            r1 = m2;
        else
            l1 = m1;
    }
    */
    for(int i = l1;i <= r1;++ i){
        int sec = need - i;
        if(l2 <= sec && sec <= r2)
            return i;
    }
    while(1);
    return l1;
}

void out(int n, int x, int from, int to){
    if(from > to)
        return;
    -- n;
    int need = x - n;
    int l1 = 0, r1 = n / 2 + (n & 1);
    while(r1 - l1 > 1){
        int mid = (l1 + r1) >> 1;
        int sec = n - mid;
        if(r[mid] + r[sec] < need)
            r1 = mid;
        else 
            l1 = mid;
    }
    if(l[l1] + l[n - l1] <= need && need <= r[l1] + r[n - l1]){
        printf("%d ", from + l1);
        int x = find_best(l[l1], r[l1], l[n - l1], r[n - l1], need);
        out(l1, x, from, from + l1 - 1);
        out(n - l1, need - x, from + l1 + 1, to);    
    }
    else {
        printf("%d ", from + r1);
        int x = find_best(l[r1], r[r1], l[n - r1], r[n - r1], need);
        out(r1, x, from, from + r1 - 1);
        out(n - r1, need - x, from + r1 + 1, to);            
    }
}

int main() {
    l[1] = r[1] = 0;
    for(int i = 2;i <= 100000;++ i){
        r[i] = i - 1 + r[i - 1];
        l[i] = i - 1 + l[(i - 1) / 2] + l[(i - 1) / 2 + (~i & 1)];
    }
    int q;
    scanf("%d", &q);
    while(q --){
        int n, x;
        scanf("%d%d", &n, &x);
        if(l[n] <= x && x <= r[n]){
            out(n, x, 1, n);
            putchar('\n');
        }
        else {
            puts("-1");
        }
    }
    return 0;
}