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 rs(): return input().strip()
def ri(): return int(input())
def ria(): return list(map(int, input().split()))
def ia_to_s(a): return ' '.join([str(s) for s in a])



def solve(n, k):
    if n < k:
        return []
    if n == k:
        return [1] * k

    # n > k

    if n % 2 == 0:
        if n >= 2*k:
            return [2] * (k-1) + [n - 2*(k-1)]
        else:
            if k % 2 == 0:
                return [1] * (k - 1) + [n - k + 1]
            else:
                return []
    else:
        #only odds
        if k % 2 == 0:
            return []
        else:
            return [1] * (k-1) + [n-k+1]


def main():
    for _ in range(ri()):
        n, k = ria()
        a = solve(n, k)
        if len(a) > 0:
            print('YES')
            print(ia_to_s(a))
        else:
            print("NO")


if __name__ == '__main__':
    main()