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 long long M = 10000000007;
const long long N = 1001;
void print(long long arr[], long long n) {
  for (long long i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
void print(int arr[], int n) {
  for (long long i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
void print(int arr[], long long n) {
  for (long long i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
void print(vector<long long> arr, long long n) {
  for (long long i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
long long power(long long a, long long n) {
  long long res = 1, mod = M;
  while (n) {
    if (n % 2) res = (res * a) % mod;
    a = (a * a) % mod;
    n = n / 2;
  }
  return res;
}
vector<long long> fact;
void calculate_factorials(long long n) {
  fact.resize(n + 1, 1);
  long long ans = 1;
  for (long long i = 1; i < n + 1; i++) {
    ans = ((ans % M) * (i % M)) % M;
    fact[i] = ans;
  }
}
void init_code() {}
void solve() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  init_code();
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v;
    if (k == 1 || k == 2) {
      v.push_back(k);
    } else {
      v.push_back(k);
      v.push_back(1);
    }
    cout << n - v.size() << endl;
    for (long long i = 1; i < n + 1; i++) {
      if (find(v.begin(), v.end(), i) == v.end()) cout << i << " ";
    }
    cout << endl;
  }
}
int main() { solve(); }
