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 parity(n,k):

	lst = []
	s = 'YES'

	if n%2 !=0 and k%2 == 0:
		return 'NO',[]

	else:
		if n%2 == 0 and k <= n//2:
			lst+= ['2']*(k-1)
			last = n-(2*(k-1))

			if last %2 != 0 or last<=0:
				return 'NO',[]
			lst.append(str(last))

		

		# elif n%2 == 0 and k > n//2 :
		# 	lst += ['1']*(k-1)
		# 	last = n-(1*(k-1))

		# 	if last %2 == 0:
		# 		return 'NO',[]
		# 	lst.append(str(last))

		else:
			lst += ['1']*(k-1)
			last = n-(1*(k-1))

			if last %2 == 0 or last<=0:
				return 'NO',[]
			lst.append(str(last))


	return s,lst

t = int(input())

for _ in range(t):
	n,k = list(map(int,input().split()))

	s,arr = parity(n,k)

	if s=='NO':
		print(s)

	else:
		print(s)
		print(" ".join(arr))