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:

n=int(input())
i=0
while i<n:
    x,y=map(int,input().split(" "))
    # print(x,y)
    if x>=y:

            # print(x-(2*(y-1)))
            if (x-(2*(y-1)))%2==0 and (x-(2*(y-1)))>0 and x!=y :
                print("YES")
                ans=[]
                ans=[2]*(y-1)
                ans=[2]*(y-1)
                ans.append(x-2*(y-1))
                print(' '.join(map(str,ans)))
            elif (x - (y - 1)) % 2 != 0 and x!=y:
                print("YES")
                ans = []
                ans = [1]*(y-1)
                ans.append(x - (y-1))
                print(' '.join(map(str,ans)))
            elif x==y:
                print("YES")
                ans=[]
                ans=[1]*y
                print(' '.join(map(str,ans)))


            else:
                print("NO")

    else:
        print("NO")

    i=i+1