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 math import *
from collections import *
from operator import itemgetter
import bisect
ii = lambda: int(input())
iia = lambda: list(map(int,input().split()))
isa = lambda: list(input().split())
I = lambda:list(map(int,input().split()))
chrIdx = lambda x: ord(x)-96
idxChr = lambda x: chr(96+x)

t = ii()
for _ in range(t):
    n,k = iia()
    if(n%2==1):
      if(n-k+1)%2==1 and (n-k+1)>0:
        print("YES")
        print('1 '*(k-1)+str(n-k+1))
      else:
        print('NO')
    elif(n%2==0):
      if((n-2*(k-1))>1) and (n-2*(k-1))%2==0:
        print("YES")
        print('2 '*(k-1)+str(n-2*(k-1)))
      elif(n-k+1)%2==1 and (n-k+1)>0:
        print("YES")
        print('1 '*(k-1)+str(n-k+1))
      else:
        print('NO')