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<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll>PLL;
typedef pair<int,int>PII;
typedef pair<double,double>PDD;
#define I_int ll
inline ll read() {
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a,ll b,ll p) {
    ll res=1;
    while(b) {
        if(b&1)res=res*a%p;
        a=a*a%p;
        b>>=1;
    }
    return res;
}
#define PI acos(-1)
const double eps=1e-8;
const int maxn=1e6+7;

int main() {
    vector<int>v;
    int t=read;
    while(t--){
        v.clear();
        int n=read,k=read;
        map<int,int>mp;
        mp[0]=1;
        for(int i=1;i<=n;i++){
            if(i==k) continue;
            if(k-i!=i&&!mp[k-i]&&k-i>=1&&k-i<=n){
                mp[i]=1;continue;
            }
            v.push_back(i);
        }
        cout<<v.size()<<endl;
        for(int x:v) cout<<x<<" ";
        if(v.size()) puts("");
    }
    return 0;
}