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:
'''#include <iostream>
#include<string.h>
#include<math.h>
#include<bits/stdc++.h>

using namespace std;





int main() {
	// your code goes here
    int x,y,l,r,i;
    scanf("%d%d%d%d",&x,&y,&l,&r);
    for(i=l;i<=r;i++)
    {
        long long int temp=(x&i)*(y&i);
        printf("%d %lld\n",i,temp);
    }
    
       
	return 0;
}
'''
import math
t=int(input())
for _ in range(0,t):
    str1=input()
    str1=str1.split()
    n=int(str1[0])
    k=int(str1[1])
    sum1=(k-1)
    sum2=(k-1)*2
    if (n-sum1)>0 and (n-sum1)%2==1:
        print("YES")
        for i in range(0,k-1):
            print(1,end=' ')
        print(n-sum1)
    elif (n-sum2)>0 and (n-sum2)%2==0:
        print("YES")
        for i in range(0,k-1):
            print(2,end=' ')
        print(n-sum2)
    else:
        print("NO")