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 odd(n, k):
    if n%2 != k%2 or k>n or (n-k)%2 != 0:
        return None
    ans = [1]*k
    ans[0] += (n-k)
    return ans

def even(n, k):
    if n%2 != 0 or 2*k > n or (n-2*k) % 2 != 0:
        return None
    ans = [2]*k
    ans[0] += (n-2*k)
    return ans

def main():
    t = int(input())
    for _ in range(t):
        n, k = list(map(int, input().split()))
        l1 = odd(n, k)
        if not l1 is None:
            print('YES')
            for x in l1:
                print(x, end = ' ')
            print()
        else:
            l2 = even(n, k)
            if not l2 is None:
                print('YES')
                for x in l2:
                    print(x, end = ' ')
                print()
            else:
                print('NO')
    
main()