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 int M = 1e9 + 7;
long long mod(long long x) { return ((x % M + M) % M); }
long long add(long long a, long long b) { return mod(mod(a) + mod(b)); }
long long mul(long long a, long long b) { return mod(mod(a) * mod(b)); }
long long modPow(int x, int y) {
  if (y == 0) return 1LL;
  if (y == 1) return mod(x);
  long long res = 1;
  while (y) {
    if (y % 2 == 1) res = mul(res, x);
    x = mul(x, x);
    y /= 2;
  }
  return res;
}
const int N = 2e5 + 2;
int fact[N];
void precalc() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
}
long long int inv(long long int x) { return modPow(x, M - 2); }
long long int divide(long long int a, long long int b) {
  return mul(a, inv(b));
}
long long int nCr(long long int n, long long int r) {
  return divide(fact[n], mul(fact[r], fact[n - r]));
}
void io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void solve() {
  int n, k;
  cin >> n >> k;
  int ans = n - (k + 1 / 2);
  cout << ans << "\n";
  for (int i = (k + 1) / 2; i <= n; i++) {
    if (i == k) continue;
    cout << i << " ";
  }
  cout << "\n";
  return;
}
int32_t main() {
  io();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
