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())

for _ in range(t):
	n,k  = map(int,input().split())
	if k > n:
		print('NO')
	elif k == 1:
		print('YES')
		print(n)
	elif n == k:
		print('YES')
		for i in range(n):	print(1,end = " ")
		print()
	else:
		# try odd:
		daru = []
		for i in range(k-1):	daru.append(1)
		if (1 + n - k)%2 == 1 and (1 + n - k) > 0:
			daru.append(1 + n - k)
			print('YES')
			for ele in daru :	print(ele,end = " ")
			print()
		else:
			daru = []
			for i in range(k-1):	daru.append(2)
			if (2 + n - 2*k)%2 == 0 and (2 + n - 2*k) > 0:
				daru.append(2 + n - 2*k)
				print('YES')
				for ele in daru :	print(ele,end = " ")
				print()
			else:
				print('NO')