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:
import sys

n = int(sys.stdin.readline())

for _ in range(n):
    n, k = sys.stdin.readline().strip().split()
    n, k = int(n), int(k)

    if n % 2 == 0:
        # even
        if n < k or (k % 2 == 1 and n < 2 * k):
            print("NO")
        else:
            print("YES")
            if k % 2 == 0:
                digits = ["1"] * (k-1) + [str(n - (k - 1))]
            else:
                digits = ["2"] * (k-1) + [str(n - 2 * (k-1))]
            print(" ".join(digits))
    else:
        # odd
        if k % 2 == 0 or k > n:
            print("NO")
        else:
            print("YES")
            digits = ["1"] * (k-1) + [str(n - (k - 1))]
            print(" ".join(digits))