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())

for i in range(t):
    n, k = [int(i) for i in input().split()]
    if k == n:
        print('YES')
        print('1 ' * n)
    elif (n - (k-1)) % 2 == 1 and (n - (k-1))> 0:
        print('YES')
        print('1 ' * (k-1)  + str(n-(k-1)))
    elif (n-2*(k-1)) % 2 == 0 and n-2*(k-1) > 1:
        print('YES')
        print('2 '*(k-1) + str(n-2*(k-1)))
    else:
        print('NO')
    # elif n % 2 == 1:
    #     if k < n:
    #         if k % 2 == 0:
    #             print('NO')
    #         else:
    #             print('YES')
    #             print('1 '*(k-1) + str(n-k+1))
    #     else:
    #         print('NO')
    # else:
    #     if k < n:
    #         if k % 2 == 0:
    #             if ((n - k + 2) // 2) % 2 != 0:
    #                 print('YES')
    #                 print('1 ' * (k-2) + str((n - (k-2))//2) + ' ' + str((n - (k-2))//2))
    #             elif 2*k <= n:
    #                 print('YES')
    #                 print('2 ' * (k - 1) +str(n-2*(k-1)))
    #             else:
    #                 print('NO')
    #         else:
    #             if 2*k <= n:
    #                 print('YES')
    #                 print('2 ' * (k-1) + str(n-(2*(k-1))))
    #             else:
    #                 print('NO')
    #     else:
    #         print('NO')