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())
ans = []
for case in range(t):
	flag = 0
	ans = []
	n,k = [int(n) for n in input().split()]
	if n < k or (n%2 == 1 and k%2 == 0):
		flag = 1
		print("NO")
	elif (n%2 == 0 and k%2 == 1):
		if n<(2*k):
			flag = 1
			print("NO")
		elif (n-k+1)%2 == 0:
			ans.append(n-2*(k-1))
			if ans[0] < 0 or ans[0] % 2 ==1:
				flag = 1
				ans[0] = 0
			for i in range(k-1):
				ans.append(2)
			if sum(ans) == n:
				print("YES")
			else:
				print("NO")
				flag = 1
	elif (n-k+1)%2 != 0:
		ans.append(n-k+1)
		for i in range(k-1):
			ans.append(1)
		if sum(ans) == n:
			print("YES")
		else:
			print("NO")
			flag = 1
	if flag == 0:
		for i in ans:
			print(i, end = " ")
		print()