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())
tf = []
c=[]
s=''
n=0
k=0
for j in range(t):
	s=input()
	n = int(s[:s.find(' ')])
	k = int(s[s.find(' ')+1:])
	if (n % 2 == 1 and k % 2 == 0) or k>n:
		tf.append(False)
		c.append([])
		continue
	elif n%2==0 and k%2==1:
		c.append([2 for i in range(k-1)])
		c[-1].append(n-2*(k-1))
	elif n%2==0 and k%2==0:
		c.append([1 for i in range(k-1)])
		c[-1].append(n-k+1)
	elif n%2==1 and k%2==1:
		c.append([1 for i in range(k-1)])
		c[-1].append(n-k+1)
	if c[-1][-1]<=0:
		tf.append(False)
		continue
	tf.append(True)
for i in range(t):
	print('YES' if tf[i] == True else 'NO')
	if tf[i] == False:
		continue
	for j in c[i]:
		print(j,end=' ')
	print('')