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:
t= int(input())
while t:
    t -= 1
    n,k = map(int, input().split())
    if n < k :
        print('NO')
    else:
        if n== k :
            ans = [1]*k
            
        elif n%2 and k%2 == 0:
            #n is odd and k is even
            print('NO')
            continue
        elif n%2 and k%2:
            # n is odd and k is odd
            ans = [1]* (k-1) + [n- k + 1]

        elif n%2 == 0 :
            if k%2:
                if 2*k > n:
                    print('NO')
                    continue
                ans = [2]* (k-1)+ [(n - (k-1)*2)]
            else:
                ans = [1]*(k-1) + [n- (k-1)]
        if len(ans) == k:
            print('YES')
            print(*ans)
        else:
            print('NO')