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--)
    {
        int n,k,a,b,c;
        cin>>n>>k;
        for(int i=0;i<k-3;i++)
        {
            cout<<"1 ";
        }
        if(k>3)
        {
            n=n-(k-3);
        }
        if(n==0)
        {
            continue;
        }
        else if(n%2==0)
        {
            if((n/2)%2==0)
            {
                a=n/2;
                c=b=(n/2)/2;
            }
            else
            {
                a=2;
                b=(n-2)/2;
                c=(n-2)/2;
            }
        }
        else
        {
            a=1;
            n=n-1;
            b=n/2;
            c=n/2;
        }
        cout<<a<<" "<<b<<" "<<c<<"\n";
    }
}