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>
int main(void) {
  using namespace std;
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int T;
  cin >> T;
  for (int case_num = 1; case_num <= T; ++case_num) {
    int64_t n, k;
    cin >> n >> k;
    vector<int64_t> A;
    if (k > n)
      ;
    else if (n % 2 == k % 2) {
      A.resize(k);
      for (auto& a : A) a = 1;
      A[k - 1] = n - k + 1;
    } else if (n % 2 == 0 && k <= n / 2) {
      A.resize(k);
      for (auto& a : A) a = 2;
      A[k - 1] = n - (2 * (k - 1));
    }
    if (A.size() > 0) {
      cout << "YES\n";
      for (auto a : A) cout << a << ' ';
    } else
      cout << "NO";
    cout << '\n';
  }
  return 0;
}