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:
#!/usr/bin/python3

# Codeforces - Round 640
# Problem B - Same Parity Summands


def read_int():
	n = int(input())
	return n


def read_ints():
	ints = [int(x) for x in input().split(" ")]
	return ints
	

#---

def solve(n,k):
	if k > n:
		return 'NO'
	
	n1 = 1
	nums = [n1] * (k-1)
	n2 = n - sum(nums)
	if n2 % 2 == 1 and n2 > 0:
		nums.append(n2)
		return [str(x) for x in nums]
	
	n1 = 2
	nums = [n1] * (k-1)
	n2 = n - sum(nums)
	if n2 % 2 == 0 and n2 > 0:
		nums.append(n2)
		return [str(x) for x in nums]
	
	return 'NO'
	


# Main
t = read_int()
for case in range(t):
	n, k = read_ints()
	sol = solve(n, k)
	
	if sol == 'NO':
		print (sol)
	else:
		print ('YES')
		print (' '.join(sol))