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:
import math
r1 = lambda : int(input());
rm = lambda : map(int,input().split());
rls = lambda : list(rm())


test = r1()
while(test):
    test-=1;
    n,k=rm();
    flag =True;
    if((n%2==0 and k==n-1 and n>2) or (n<k) or (n%2==0 and k%2==1 and n<(2*k) )):
        flag = False;
        
    if(k%2==0 and n%2==1):
        flag = False;
    x=k-1;
    ans=[];
    if(n%2==0 and k%2==1):
        m=k-1;
        ans=[2]*m;
        x=n-(2*m);
        ans.append(x);
    elif(n%2==1 or k%2==1 or (n%2==0 and k%2==0 and n<(2*k))):
        ans=[1]*x;
        x=n-x;
        ans.append(x);
    elif(n%2==0 and k%2==0):
        m=k-1;
        ans=[2]*m;
        x=n-(2*m);
        ans.append(x);
    
        
    if(flag):
        print("YES");
        print(*ans);
    else:
        print("NO")