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:
from sys import stdin, stdout
cin = stdin.readline
cout = stdout.write
mp = lambda: list(map(int, cin().split()))

def chars(): #function for taking string input as character array since string in python is immutable
    s = cin()
    return(list(s[:len(s) - 1]))
	
def pl(a): #print list	
	for val in a:
		cout(str(val) + ' ')
	cout('\n')
	
	
	
t, = mp()
for _ in range(t):
	n, k = mp()
	
	a = []
	if not n%2:
		if k*2 > n:
			if not k%2 and k <= n:
				cout('yes\n')
				pl([1]*(k-1) + [n-k+1])
			else:
				cout('no\n')
		else:
			cout('yes\n')
			pl([2]*(k-1)+ [n-2*(k-1)])
			#a += [2]*(k-1) + [n-2*(k-1)]
	else:
		if not k%2:
			cout('no\n')
		else:
			if k > n:
				cout('no\n')
			else:	
				cout('yes\n')
				pl([1]*(k-1) + [n-k+1])