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 itertools import *
from bisect import *
def io():
    return map(int,input().split())
def op():
    return list(map(int,input().split()))
def kl(con,x=0):
    if x==0:print('Yes') if con else print('No')
    elif x==1:print('yes') if con else print('no')
    elif x==2:print('YES') if con else print('NO')
MOD = 1000000007
MAX=2147483647
MIN=-2147483648
p=input
for _ in range(int(p())):
    l=[]
    a,b=io()
    if a&1 and not b&1:
        kl(0,2)
        continue
    if not a&1 and b&1:
        rem=a-(b-1)*2
        l=[2]*(b-1)
        if rem<=0 or rem&1:
            kl(0,2)
            continue
        else:l.append(rem)
    else:
        rem=a-(b-1)
        l=[1]*(b-1)
        if rem <=0 or (not rem&1):
            kl(0, 2)
            continue
        else:
            l.append(rem)
    kl(1,2)
    print(*l)