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:
#x,y = map(int,input().split())
#arr = [int(i) for i in input().split()]
from collections import Counter
import math
mod = 10**9+7
 
for i in range(int(input())):
	
	n,k = map(int,input().split())
	#arr = [int(i) for i in input().split()]
	ans = []
	flag = 0
	if n%k == 0:
		if n<k:
			print("NO")
		else:
			x = n//k
			l = n//x
			for i in range(l):
				ans.append(x)
			print("YES")
			print(*ans)
	else:
		#n -= (k-1)
		if (n - (k-1)) %2 != 0 and (n-(k-1)) > 0:
			for i in range(k-1):
				ans.append(1)
			ans.append(n-(k-1))
			print("YES")
			print(*ans)
		elif (n-(k-1)*2) % 2 == 0 and n-(k-1)*2 > 0:
			for i in range(k-1):
				ans.append(2)
			ans.append(n-(k-1)*2)
			print("YES")
			print(*ans)
		else:
			print("NO")