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())
n = [0]*T
k = [0]*T
 
for t in range(T):
  n[t], k[t] = [int(i) for i in input().split(' ')]
 
 
def ans(n,k):
  if n%2 == 1 and k%2 ==0:
    return 'NO'
  elif k>n:
    return 'NO'
  elif n%2 == 0 and k%2 == 1 and 2*k>n:
    return 'NO'
  elif n%2 == 0 and k%2 == 1:
    return [str(2)]*(k-1)+[str(n-(2*k-2))]
  elif (n%2 == 0 and k%2==0) or (n%2==1 and k%2==1):
    return [str(1)]*(k-1)+[str(n-(k-1))]


for t in range(T):
  a = ans(n[t],k[t])
  if a == 'NO':
    print('NO')
  else:
    print('YES')
    print(' '.join(a))