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:
def solve():
    N, S = [int(el) for el in input().split()]
    can = False
    ans = []
    if N % 2 == 1:
        # Odd
        if (N - (S - 1)) > 0 and (N - (S - 1)) % 2 == 1:
            can = True
            ans = [1 for _ in range(S - 1)]
            ans.append(N - (S - 1))
    else:
        # ODD
        if (N - (S - 1)) > 0 and (N - (S - 1)) % 2 == 1:
            can = True
            ans = [1 for _ in range(S - 1)]
            ans.append(N - (S - 1))
        # EVEN
        if (N - 2 * (S - 1)) > 0 and (N - 2 * (S - 1)) % 2 == 0:
            can = True
            ans = [2 for _ in range(S - 1)]
            ans.append(N - 2 * (S - 1))
    if can:
        print("YES")
        print(" ".join(map(str, ans)))
    else:
        print("NO")


for t in range(1, int(input()) + 1):
    solve()