CODECHEF Permutation GCD Problem Solution

//Divyansh Vinayak
#include<bits/stdc++.h>
using namespace std;
#define gc getchar_unlocked
#define fo(i,n) for(i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define ll long long
#define si(x)   scanf("%d",&x)
#define sl(x)   scanf("%lld",&x)
#define ss(s)   scanf("%s",s)
#define pi(x)   printf("%d",x)
#define pl(x)   printf("%lld",x)
#define ps(s)   printf("%s",s)
#define pnl()   printf("n")
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
typedef pair<int, int>  pii;
typedef pair<ll, ll>    pl;
typedef vector<int>     vi;
typedef vector<ll>      vl;
typedef vector<pii>     vpii;
typedef vector<pl>      vpl;
typedef vector<vi>      vvi;
typedef vector<vl>      vvl;
int mpow(int base, int exp); 
void ipgraph(int m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 18 * 103, M = N;
#define top(a,b) (ll)((a+b-1)/b)
//=======================
vi g[N];
void matrix(ll n, ll m){
  char s[101][101];
    while(cin>>n>>m){
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cin>>s[i][j];
            }
        }
}
}

int fact(ll n);
 

 
// Returns factorial of n
int fact(ll n)
{
    ll res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}

int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

int main() {
  ll t;
  cin >> t;
  while(t--){
     ll a,b;
     cin >> a >> b;
     if (b<a){
         cout<<-1<<endl;
         continue;
     }
     cout<<b-a+1<<" ";
     for(int i=1;i<=a;i++){
         if (i!= b-a+1   ){
             cout<<i<<" ";
         }
     }
     cout<<endl;
  }
  return 0;
}      


int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((ll)result * base) % mod;
    base = ((ll)base * base) % mod;
    exp >>= 1;
  }
  return result;
}

void ipgraph(int n, int m){
    int i, u, v;
    while(m--){
        cin>>u>>v;
        g[u-1].pb(v-1);
        g[v-1].pb(u-1);
    }
}

void dfs(int u, int par){
    for(int v:g[u]){
        if (v == par) continue;
        dfs(v, u);
    }
}