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:
for test in range(int(input())):
    n,k = map(int,input().split())
    lt=[]
    if n/k==n//k:
        print('YES')
        for i in range(k):
            lt.append(n//k)
        print(*lt,sep=' ')
    elif n==k:
        print('YES')
        for i in range(k):
            lt.append(1)
        print(*lt,sep=' ')
    elif (n-(2*(k-1)))%2==0 and (n-(2*(k-1)))>0:
        print('YES')
        for i in range(k-1):
            lt.append(2)
        lt.append((n-(2*(k-1))))
        print(*lt,sep=' ')

    elif (n-(k-1))%2!=0 and (n-k-1)>0:
        print('YES')
        for i in range(k-1):
            lt.append(1)
        lt.append(n-k+1)
        print(*lt,sep=' ')

    else:
        print('NO')