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 
#include <bits/stdc++.h>

using namespace std;


int main(){
    int t;
    cin>>t;
    while(t--){
        long long int n,k;
        cin>>n>>k;
        
        long long int x = n;
        long long int i=1;
        while(k>3){
            n--;
            k--;
            cout<<1<<" ";
        }
        
         if(n%3==0){
             cout<<n/3<<" "<<n/3<<" "<<n/3<<endl;
         }
         else{
             if(n%2==0){
                 if((n/2)%2!=0){
                 cout<<2<<" "<<(n-2)/2<<" "<<(n-2)/2<<endl;
                 }
                 else{
                     long long int y = n/2;
                     cout<<y/2<<" "<<y/2<<" "<<y<<endl;
                 }
             }
             else{
                 cout<<1<<" "<<(n-1)/2<<" "<<(n-1)/2<<endl;
             }
         }
    }
    
}