It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.

You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:

  * a_1 + a_2 + … + a_k = n 
  * LCM(a_1, a_2, …, a_k) ≤ n/2 



Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.

We can show that for given constraints the answer always exists.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).

It is guaranteed that the sum of k over all test cases does not exceed 10^5.

Output

For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.

Example

Input


2
6 4
9 5


Output


1 2 2 1 
1 3 3 1 1 
import java.io.*;

import java.util.*;
 public class Solution {
	 public static void main(String[] args) throws Exception
    {
		 FastReader fr=new FastReader();
		 int t=fr.nextInt();
		 while(t-->0) {
		    int n=fr.nextInt();
		    int k=fr.nextInt();
		    n=n-(k-3);
		    if(n%2==0) {
		    	int v=n/2;
		    	if(v%2==0) {
		    		System.out.print(v+" "+(v/2)+" "+(v/2)+" ");
		    	}
		    	else {
		    		int f=n-2;
		    	    System.out.print(f/2+" "+(f/2)+" "+2+" ");	
		    	}
		    }
		    else {
		    	int f=n/2;
		    	System.out.print(f+" "+f+" "+1+" ");
	    	
		    }
		    
		    	k-=3;
		    	while(k-->0) {
		    		System.out.print(1+" ");
		    	}
		    
		 
		System.out.println();
		 
		 }  
	 }
	 public static boolean isPali(String s) {
		 int i=0;
		 int j=s.length()-1;
		 while(i<=j) {
			 if(s.charAt(i)!=s.charAt(j))
				 return false;
			 i++;
			 j--;
		 }
		 return true;
	 }
	 public static String reverse(String s) {
		 String res="";
		 for(int i=0;i<s.length();i++) {
			 res+=s.charAt(i);
		 }
		 return res;
	 }
	
		
	 
	 public static int bsearch(long suf[],long val) {
		 int i=0;
		 int j=suf.length-1;
		 while(i<=j) {
			 int mid=(i+j)/2;
			 if(suf[mid]==val)
				 return mid;
			 else if(suf[mid]<val)
				 j=mid-1;
			 else
				 i=mid+1;
		 }
		 return -1;
	 }
	 
	 
	 
	
	 
	 public static void print(String str) {
		 System.out.println(str);
	 }
	public static int[] getFreq(String s) {
		int a[]=new int[26];
		for(int i=0;i<s.length();i++) {
			a[s.charAt(i)-'a']++;
		}
		return a;
		
	}
	 
	 public static boolean isPrime(int n) {
		 
		 for(int i=2;(i*i)<=n;i++) {
			 if(n%i==0)
				 return false;
		 }
		 return true;
	 }
            	  
             
             
    }
 class Pair{
	 int x;
	 int y;
	 long cost;
	 Pair(int x,int y,long cost){
		 this.x=x;
		 this.y=y;
		 this.cost=cost;
	 }
 }
	
	 
	
	

	 
 
  

class FastReader
{
    BufferedReader br;
    StringTokenizer st;
 
    public FastReader()
    {
        br = new BufferedReader(new
                InputStreamReader(System.in));
    }
 
    String next()
    {
        while (st == null || !st.hasMoreElements())
        {
            try
            {
                st = new StringTokenizer(br.readLine());
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
 
    int nextInt()
    {
        return Integer.parseInt(next());
    }
 
    long nextLong()
    {
        return Long.parseLong(next());
    }
 
    double nextDouble()
    {
        return Double.parseDouble(next());
    }
 
    String nextLine()
    {
        String str = "";
        try
        {
            str = br.readLine();
        }
        catch (IOException e)
        {
            e.printStackTrace();
        }
        return str;
    }
}

