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:
for _ in range(int(input())):
	n,k=map(int,input().split())

	if k>n:
		print('No')
	elif n%2!=0 and k%2==0:
		print('No')
	elif n%2==0 and k%2==0:
		if n/k==n//k:
			a=[n//k]*k
		else:
			a=[n//k]*(k-1)
			a.append(int(n-n//k*(k-1)))
		print('Yes')
		print(*a)
	elif n%2==0 and n//k==1:
		print('No')
	else:
		if n/k==n//k:
			a=[n//k]*k
		elif n%2==(n//k)%2:
			a=[n//k]*(k-1)
			a.append(int(n-n//k*(k-1)))
		else:
			a=[n//k-1]*(k-1)
			a.append(int(n-(n//k-1)*(k-1)))
		print('Yes')
		print(*a)