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 _ in range(t):
    n,k=map(int,input().split())
    even,odd=[],[]
    o,e=n,n
    for i in range(1,k+1):
        if e>=2 and i<k:
            even.append(2)
            e-=2
        if i==k and e>0:
            even.append(e)
    for i in range(1,k+1):
        if o>=1 and i<k:
            odd.append(1)
            o-=1
        if i==k and o>0:
            odd.append(o)
    odd_=len(odd)
    even_=len(even)
    if (odd[odd_-1]&1 and odd_==k):
        print("YES")
        for i in range(odd_):
            print(odd[i],end=" ")
    elif (even[even_-1]%2==0 and even_==k):
        print("YES")
        for i in range(even_):
            print(even[i],end=" ")
    else:
        print("NO")
    print("\n")