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())
inp = []
    
for i in range(n):
    inp.append([])
    n1, n2 = map(int, input().split(' '))
    inp[i].append(n1)
    inp[i].append(n2)

def preds(a, b):
    if a < b:
        return('NO')

    if (a-b+1) % 2 == 1:
        ret = []
        for i in range(b-1):
            ret.append('1')
        ret.append(str(a-b+1))
        return(ret)
    else:
        if a < 2*b:
            return('NO')
        
        if ((a-b) - (b-1)) % 2 == 1:
            ret = []
            for i in range(b-1):
                ret.append('2')
            ret.append(str(a - (2*(b-1))))
            return(ret)
        else:
            return('NO')

for i in range(len(inp)):
    ans = preds(inp[i][0], inp[i][1])
    if ans == 'NO':
        print('NO')
    else:
        print('YES')
        print(' '.join(ans))