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, exit
input = stdin.readline

def i(): return input()
def ii(): return int(input())
def iis(): return map(int, input().split())
def liis():	return list(map(int, input().split()))
def print_array(a): print(" ".join(map(str, a)))

t = ii()
for _ in range(t):
	n, k = iis()
	if (n-(2*(k-1)))%2 == 0 and (n-(2*(k-1))) > 0:
		print("YES")
		a = []
		for i in range(k-1):
			a.append(2)	
		a.append(n-sum(a))
		print_array(a)
	elif (n-(1*(k-1)))%2 == 1 and (n-(1*(k-1))) > 0:
		print("YES")
		a = []
		for i in range(k-1):
			a.append(1)	
		a.append(n-sum(a))
		print_array(a)
	else:
		print("NO")