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>
#include<cstring>
#include<climits>
using namespace std;
#define add(x) accumulate(x)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rep(i, a, b) for (int i=a; i<(b); i++)
#define hs ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
const ll mod = 1000000007;
const ll INF = 1e18;
const ll MAX = 100001;

void solve() {
	// code goes here
	ll n, k;
	cin >> n >> k;
	vector<ll> vec;
	for (int i = k + 1; i <= n; i++) {
		vec.push_back(i);
	}
	map<ll, bool> mp;
	for (int i = 1; i < k; i++) {
		mp[i] = false;
	}
	for (int i = k - 1; i >= 1; i--) {
		if (mp[i] == false && mp[k - i] == false) {
			vec.push_back(i);
			mp[i] = true;
			mp[k - i] = true;
		}
	}
	cout << vec.size() << endl;
	for (int i = 0; i < vec.size(); i++) {
		cout << vec[i] << " ";
	}
	cout << endl;
	return;
}
int main() {


	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.precision(12);
	cout << fixed;


	ll t;
	cin >> t;
	while (t--) {
		solve();
	}
	return 0;
}