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;
    scanf("%d",&t);
    while(t--){
        int n,k;
        scanf("%d%d",&n,&k);
        int tmp=n-k+3;
        for(int i=1;i<=k-3;i++){
            printf("1 ");
        }
        if(tmp%2==1){
            printf("1 %d %d\n",tmp/2,tmp/2);
        }else{
            int x=1;
            while(tmp%2==0){
                x<<=1;
                tmp>>=1;
            }
            if(tmp!=1)
                printf("%d %d %d\n",x,x*((tmp-1)/2),x*((tmp-1)/2));
            else
                printf("%d %d %d\n",x/4,x/4,x/2);
        }
    }
}