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=[]
k=[]
for i in range(t):
    inpn, inpk=map(int,input().split())

    n.append(inpn)
    k.append(inpk)



for i in range(t):

 l = (n[i] // 2)
 j = k[i]

 if(n[i]%2==0 and l//j!=0):

     print('YES')
     rem=-1
     while(rem!=0):
         rem=(l%j)
        # print(l, j, rem)
         if(rem!=0):
           print((2*((l)%j)),end=' ')

         else:
             print(((str(((2*l//j)))+' ')*j))


         (l) = l - rem
         j=j-1

 elif(n[i]-k[i]>=0 and (n[i]-k[i])%2==0):
     l=((n[i]-k[i])//2)
     j=k[i]
     print('YES')
     rem=-1
     while(rem!=0):
         rem=(l%j)
        # print(l, j, rem)
         if(rem!=0):
           print((2*((l)%j)+1),end=' ')

         else:
             print(((str(((2*l//j)+1))+' ')*j))


         (l) = l - rem
         j=j-1


 else:
     print('NO')