Problem: You are given two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).

In other words, find a_1, a_2, …, a_k such that all a_i>0, n = a_1 + a_2 + … + a_k and either all a_i are even or all a_i are odd at the same time.

If such a representation does not exist, then report it.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100).

Output

For each test case print:

  * YES and the required values a_i, if the answer exists (if there are several answers, print any of them); 
  * NO if the answer does not exist. 



The letters in the words YES and NO can be printed in any case.

Example

Input


8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9


Output


YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
void solve() {
  long long int n, k;
  cin >> n >> k;
  bool f = false;
  if ((n - k + 1) % 2 != 0 and (n - k + 1) > 0) {
    f = true;
    cout << "YES" << endl;
    for (long long int i = 0; i < (k - 1); ++i) {
      cout << 1 << " ";
    }
    cout << (n - k + 1) << endl;
  }
  if (f == false and (n - 2 * k + 2) > 0 and (n - 2 * k + 2) % 2 == 0) {
    f = true;
    cout << "YES" << endl;
    for (long long int i = 0; i < (k - 1); ++i) {
      cout << 2 << " ";
    }
    cout << (n - 2 * k + 2) << endl;
  }
  if (f == false) {
    cout << "NO" << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tc;
  cin >> tc;
  for (long long int i = 1; i <= tc; ++i) {
    solve();
  }
  return 0;
}