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 fun(n,k):
    if k>n:
        return "NO"
    if k%2==0 and n%2==1:
        return "NO"
   
    if k%2==1:
        if n%2==1: # odd parity
           ans=([1]*(k-1))+[n-(k-1)]
        else:
           if 2*(k-1)>=n:
              return "NO"
           ans=([2]*(k-1))+[n-2*(k-1)]   
    
    else: # k is even
        if n%2==1:
           return "NO"
        else: # n is even
             ans=([1]*(k-1))+[n-(k-1)]   

    return ans



for _ in range(int(input())):
    a,b=[int(x) for x in input().split()]
    x=fun(a,b)
    if x=="NO":
        print(x)
    else:
      print("YES")
      print(" ".join([str(i) for i in x]))