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
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define ff first
#define ss second
#define pii pair <int,int>
#define pll pair <ll,ll>
#define testcase() int t; cin>>t; while(t--)
#define forn(i,n) for(int i=0;i<n;i++)
#define forn1(i,n) for(int i=1;i<=n;i++)
#define vll vector <ll>
#define vi vector <int>
#define all(v) v.begin(),v.end()

ll M = 1e9 + 7;
double pi = acos(-1.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll powerm(ll x,ll y){ ll res=1; while(y){ if(y&1) res=(res*x)%M; y=y>>1; x=(x*x)%M;} return res%M; }
ll power(ll x,ll y){ ll res=1; while(y){ if(y&1) res=(res*x); y=y>>1; x=(x*x);} return res; }
ll gcd(ll a,ll b){if(b>a)return gcd(b,a); if(b==0)return a; return gcd(b,a%b);}

int main() {
	
	//ifstream cin("input.txt");
	//ofstream cout("output.txt");
	ios_base::sync_with_stdio(false); 
    cin.tie(NULL); 
	
	testcase()
	{
		int n,k; cin>>n>>k;
		int ans = k/2 + n - k;
		cout<<ans<<'\n';
		int st = (k+1)/2;
		while(st<k)
		{
			cout<<st<<' ';
			st++;
		}
		for(int i=k+1;i<=n;i++) cout<<i<<' ';
		cout<<'\n';

	}

	return 0;
}
