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())
ans = []
for s in range(t):
    n, k = [int(x) for x in input().split()]
    if (((k * 1) % 2 != n % 2) and ((k * 2) % 2 != n % 2)) or k > n:
        ans.append([-1])
    else:
        if (k * 1) % 2 == n % 2:
            arr = [1 for i in range(k-1)]
            p = sum(arr)
            if p>=n:
                ans.append([-1])
            elif p != n:
                arr.append(n-p)
                ans.append(arr)
        elif (k * 2) % 2 == n % 2:
            arr = [2 for i in range(k-1)]
            p = sum(arr)
            if p>=n:
                ans.append([-1])
            elif p != n:
                arr.append(n-p)
                ans.append(arr)
m=0
while m<len(ans):
    if ans[m]==[-1]:
        print('NO')
        m+=1
    else:
        print('YES')
        print(*ans[m])
        m+=1