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:
# 640, Div 4
# B. Same Parity Summands
import sys

def solve():
	return None

def main():
	q = []
	for line in sys.stdin.readlines():
		q.append(line)

	for i in range(len(q)):
		q[i] = q[i].rstrip().split(' ')
		q[i] = [int(x) for x in q[i]]

	for i in range(1, len(q)):
		# print(q[i])
		n = q[i][0]
		k = q[i][1]
		# solve(q[i])
		# easy win
		# if n%k == 0:
		# 	print("yes")
		# 	top = str(int(n/k)) + " "
		# 	print(top * k)
		# else if 
		# always try 1 or 2 and see last number
		oddtry = n
		eventry = n
		for i in range(k - 1):
			oddtry -= 1
			eventry -= 2
		if eventry % 2 == 0 and eventry >= 2:
				print("YES")
				print("2 " * (k-1), end = '')
				print(str(eventry))
		elif oddtry % 2 == 1 and oddtry >= 1:
			print("YES")
			print("1 " * (k-1), end = '')
			print(str(oddtry))
		else:
			print ("NO")
			


if __name__ == '__main__':
	main()