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:
  [n,k] = [int(x) for x in input().split()]

  # see if there are k even summands for n
  # take k-1 1s then remaining n-2(k-1) should be > 0 and even
  if n - 2*(k-1) > 0 and (n - 2*(k-1))%2 == 0:
    print('yes')
    ans = ''
    for i in range(k-1):
      ans += '2 '
    ans += str(n-2*k+2)
    print(ans)

  # see if there are k odd summands for n
  # take k-1, 2s then the remaining n-2(k-1) should be > 0 and even
  elif n - (k-1) > 0 and (n - (k-1))%2 != 0:
    print('yes')
    ans = ''
    for i in range(k-1):
      ans += '1 '
    ans += str(n-k+1)
    print(ans)
  
  else: print('no')

  t-=1