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;
long long M = 1e9 + 7;
long long mm(long long x, long long y) {
  x %= M, y %= M;
  return (x * y) % M;
}
long long po(long long x, long long y) {
  if (!y) return 1;
  long long a = po(x, y / 2) % M;
  if (y % 2) return mm(a, mm(a, x));
  return mm(a, a);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, c = 0, t = 1, i, j, k, sum = 0, ans, f = 0, x, y;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    if (n >= m and (n - m) % 2 == 0) {
      cout << "YES\n";
      for (long long i = 1; i < m; i++) cout << "1 ";
      cout << n - m + 1;
    } else if (n >= 2 * m and (n - 2 * m) % 2 == 0) {
      cout << "YES\n";
      for (long long i = 1; i < m; i++) cout << "2 ";
      cout << n - 2 * m + 2;
    } else
      cout << "NO";
    cout << endl;
  }
  return 0;
}