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 PYTHON3):

Solution:
def find_terms(n, k):
    if k == 1:
        return '\n'.join(["YES", str(n)])
    if k > n or n % 2 != 0 and k % 2 == 0:
        return "NO"
    if k == n:
        return '\n'.join(["YES", ' '.join([str(1)]*n)])
    if n % 2 == 0 and k % 2 == 0 or n % 2 != 0 and k % 2 != 0:
        res = [str(1)]*(k-1) + [str(n-k+1)]
        return '\n'.join(["YES", ' '.join(res)])
    if n % 2 == 0 and k*2 <= n:
        res = [str(2)]*(k-1) + [str(n - 2*(k-1))]
        return '\n'.join(["YES", ' '.join(res)])
    return "NO"


if __name__ == '__main__':
    pairs = []
    for _ in range(int(input())):
        pairs.append([int(s) for s in input().split()])
    results = [find_terms(n, k) for n, k in pairs]
    print('\n'.join(results), sep='')