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:
#include<iostream>
#include <cstddef>
#include<set>
#include<stack>
#include <deque>
#include<queue>
#include<utility>
#include<vector>
#include<map>
#include<cmath>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<iomanip>
#include<algorithm>
#define  ll  long long
#define ull unsigned long long
#define  sl(a)  scanf("%lld",&a)
#define  si(a)  scanf("%d",&a)
#define mem(n,m) memset(n,m,sizeof(n))
#define PI acos(-1)
#define pb    push_back
#define pf    push_front
#define vll   vector<ll>
#define pll   pair<ll,ll>
#define mp  make_pair
#define vpll vector< pair <ll,ll> >
#define vplpll vector<pair<ll,pair<ll,ll> > >
#define mp make_pair
#define all(v) v.begin(),v.end()
#define S(X) ( (X) * (X) )
#define forab(i, a, b) for(i = a; i <= b; i++)
#define forn(i, n) for(i = 0; i < n; i++)
#define print(a,b) printf("Case %lld: %lld\n",a,b)
#define f first
#define s second
using namespace std;
ll powd(ll a,ll b){if(b==1){return a;}if(b%2==1){return powd(a,b-1)*a;}else{ll x=powd(a,b/2); return x*x;}}
ll powm(ll a,ll b,ll m){if(b==0){return 1;}if(b==1){return a%m;}if(b%2==1){return ((powm(a,b-1,m)%m)*(a%m))%m;}else{ll x=powm(a,b/2,m); return ((x%m)*(x%m))%m;}}
ll gcd(ll a,ll b){if(a==0){return b;}return gcd(b,a%b);}
int main()
{
  ll n,k;
  ll t;
  cin>>t;
  while(t--)
  {
      sl(n);
      sl(k);
      int ans=(n-k)+(k/2);
      cout<<ans<<endl;
      for(int i=k+1;i<=n;i++)
      {
          cout<<i<<" ";
      }
      int u=k/2;
      if(k%2==1)
      {
          u++;
      }
      for(int i=u;i<k;i++)
      {
          cout<<i<<" ";
      }
      cout<<endl;
  }
  return 0;
}