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 i in range(t):
    n,k = input().split()
    n,k = int(n),int(k)
    if k == 1:
        print("YES")
        print(n)
    elif (n%2) == (k%2):
        if (n-k+1) <= 0:
            print("NO")
        else:
            print("YES")
            e = ['1']*(k-1)
            e.append(str(n-k+1))
            print(" ".join(e))
    elif ((n%2) == 0) and (n >= (2*k)):
        if (n - (2*(k-1))) <= 0:
            print("NO")
        else:
            print("YES")
            e = ['2'] * (k - 1)
            e.append(str(n - (2*(k-1))))
            print(" ".join(e))
    else:
        print("NO")