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>
#define int long long
using namespace std;
int32_t main() {
    int t;
    cin>>t;
    while(t--)
    {
        int n,k;
        cin>>n>>k;
        int temp=n-(k-3);
        if(temp%2==0)
        {
            int num=temp/2;
            if(num%2==0)
            {
                int half=num/2;
                cout<<num<<" "<<half<<" "<<half<<" ";
            }
            else
            {
                int half=num-1;
                cout<<half<<" "<<half<<" "<<2<<" ";
            }
        }
        else
        {
            int num=temp/2;
            cout<<num<<" "<<num<<" "<<1<<" ";
        }
        for(int i=0;i<k-3;i++) cout<<1<<" ";
        cout<<endl;
    }
}