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 sys
# sys.setrecursionlimit(10**6) 
input=sys.stdin.readline
t=int(input())

for i1 in range(t):
    n,k=list(map(int,input().split(" ")))
    
    arr=[0]*k
    fl=0
    for i in range(k):
        arr[i]=n//k
        if(i==k-1):
            
            arr[i]+=n%k
            
    z="YES"
    # print(fl)
    if(len(arr)>1):
        if(arr[-1]%2!=arr[-2]%2):
            fl=0
            for i in range(0,k-1,2):
                arr[i]+=1
                arr[i+1]-=1
                if(arr[i+1]==0):
                    fl=1
        if(arr[-1]%2!=arr[-2]%2 or fl==1 or k>n):
            print("NO")

        else:
            print("YES")
            s=""
            for i in range(k):
                s+=str(arr[i])+" "
            print(s)



    else:
        print(z)
        print(arr[0])