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:
c = int(input())
n = []
for i in range(c):
	n.append(list(map(int, input().split())))

def find_pairs(x):
	val = x[0]
	cou = x[1]
	i = cou
	r = []
	if (val % 2) == (cou % 2):
		# odd nums	
		while (i != 1):
			r.append(1)
			i -= 1
		t = val - sum(r)
		if t < 1:
			return False
		if t % 2 == 1:
			r.append(t)
			return r
		else :
			return False
	else :
		# even nums
		while (i != 1):
			r.append(2)
			i -= 1
		t = val - sum(r)
		if t < 1:
			return False
		if t % 2 == 0:
			r.append(t)
			return r
		else :
			return False

for i in n:
	fp = find_pairs(i)
	if fp:
		print("YES")
		print(' '.join(list(map(str, fp))))
	else :
		print("NO")