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 func(x):
    x = x.split()
    if len(x) == 1:
        return int(x[0])
    x = [int(each) for each in x]
    return x
 
import sys 
inputs = []
for line in sys.stdin: 
    inputs += [line.split('\n')[0]]
inputs = [func(each) for each in inputs]
for i in range(inputs[0]):
    n, k = inputs[i+1][0], inputs[i+1][1]
    #Check if even possible
    last_even = n-(2*k - 2)
    if (last_even > 0) & (last_even%2 == 0):
        ans = [2]*(k-1) + [last_even]
        ans = [str(each) for each in ans]
        print("YES")
        print(' '.join(ans))
        continue
    last_odd = n-k+1
    if (last_odd > 0) & (last_odd%2 != 0):
        ans = [1]*(k-1) + [last_odd]
        ans = [str(each) for each in ans]
        print("YES")
        print(' '.join(ans))
        continue
    
    print("No")