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;
const int MAX_N = 1e5 + 1;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long p = k / 2;
  long long ans = p + n - k;
  if (n == 1) {
    cout << "0" << endl;
    return;
  }
  if (k >= (2 * n)) {
    cout << n << endl;
    for (long long i = 1; i < n; i++) {
      cout << i << " ";
    }
    cout << n << endl;
    return;
  }
  vector<long long> v;
  for (long long i = 1; i <= p; i++) {
    v.push_back(i);
  }
  for (long long i = k + 1; i <= n; i++) {
    v.push_back(i);
  }
  cout << v.size() << endl;
  for (long long i = 0; i < v.size() - 1; i++) {
    cout << v[i] << " ";
  }
  cout << v[v.size() - 1] << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
}
