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 N = 1000010;
vector<long long> is_prime(N + 1, true);
long long to_int(string s) {
  stringstream geek(s);
  long long x = 0;
  geek >> x;
  return x;
}
string to_string(long long n) {
  ostringstream str1;
  str1 << n;
  string geek = str1.str();
  return geek;
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long powerm(long long a, long long k) {
  if (!k) {
    return 1;
  }
  long long b = powerm(a, k / 2);
  b = b * b % (1000000000 + 7);
  if (k % 2) {
    return a * b % (1000000000 + 7);
  } else {
    return b;
  }
}
long long power(long long a, long long b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long m1 = power(a, b / 2);
  if (b % 2) return m1 * m1 * a;
  return m1 * m1;
}
void seive_of_eras(long long n) {
  is_prime[0] = 0;
  is_prime[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
bool isprime(long long a) {
  if (a <= 1) return false;
  if (a == 2 || a == 3) return true;
  if (a % 2 == 0 || a % 3 == 0) return false;
  for (long long i = 5; i * i <= a; i = i + 6) {
    if (a % i == 0 || a % (i + 2) == 0) return false;
  }
  return true;
}
void codeit() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v;
  long long s = 0;
  long long p = -1;
  for (long long i = 1; i < k; i++) {
    s = s + i;
    if (s >= k) {
      p = i;
      break;
    }
  }
  if (k == 2) {
    v.push_back(1);
  }
  if (p <= 0) p = k;
  for (long long i = p; i < k; 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(); i++) {
    cout << v[i] << " ";
  }
  cout << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long tt;
  tt = 1;
  cin >> tt;
  for (long long i = 0; i < tt; i++) {
    codeit();
  }
}
