Problem: You are given two integers n and k. You are asked to choose maximum number of distinct integers from 1 to n so that there is no subset of chosen numbers with sum equal to k.

A subset of a set is a set that can be obtained from initial one by removing some (possibly all or none) elements of it.

Input

The first line contains the number of test cases T (1 ≤ T ≤ 100).

Each of the next T lines contains two integers n and k (1 ≤ k ≤ n ≤ 1000) — the description of test cases.

Output

For each test case output two lines. In the first line output a single integer m — the number of chosen integers.

In the second line output m distinct integers from 1 to n — the chosen numbers.

If there are multiple answers, print any. You can print the numbers in any order.

Example

Input


3
3 2
5 3
1 1


Output


2
3 1 
3
4 5 2 
0

Solution (in CPP):

Solution:
// Premature optimization is the root of all evil in programming - Knuth.

#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;

#define mod (int) (1e9+7)

ll gcd(ll a, ll b){
	
	if(b==0){

	   return a;
	}
	
	return gcd(b, a%b);
	
}
 
 ll ncr(ll x, ll y){
 	ll ans=1;
 	
 	y=(y > x-y? x-y: y);
 	
 	for(ll i=0; i<y; i++){

 		ans=ans*(x-i);

 		ans=ans/(i+1);

 		ans=ans%mod;
 	}
 
 	
 	return ans;
 }

ll power(ll x, ll y){

    ll ans=1;

    while(y){

        if(y&1){

            ans=ans*x;

            ans=ans%mod;

        }
        
        x=x*x;

        x=x%mod;

        y=y>>1;
    }
    
    return ans;
    
}
 
void solve(){
	
	ll n,k; cin>>n>>k;
	
	ll res=0; res+=0;
	
	set<ll> s; res+=(k)/2;
	
	for(int i=k-1; i>=k-res; i--){
		// ll val=(i*(i+1)); val=val/2;
		// if(val<k){
			// res++;
			s.insert(i);
		// }
	}
	
	for(int i=k+1; i<=n; i++){
		s.insert(i);
	}
	
	res+=n-k;
	
	cout<<res<<"\n";
	
	for(auto it: s){
		cout<<it<<" ";
	}cout<<"\n";
	
	return;
	
}


int main(){
    ios_base::sync_with_stdio(0);
    
    cin.tie(0);
    
    int t=1;cin>>t;
   
    // clock_t begin, end;
    
    // double time_spent;
    
    // begin = clock(); // Time taken till now.
    
    while(t--){
    
	  solve();
        
    }
    
   //end= clock(); // Total  Time taken till now.
   
   //time_spent= (end- begin) / CLOCKS_PER_SEC; 
   
   //cout<<time_spent<<"\n";
   
	return 0;
}