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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

#define lowbit(x) ((x)&(-(x)))

using namespace std;

long long n, k;
int t;

int main()
{
	scanf("%d", &t);
	while (t--)
	{
		scanf("%lld%lld", &n, &k);
		while (k > 3)
		{
			printf("1 ");
			--k;
			--n;
		}
		if (n % 2)
		{
			printf("%lld %lld %lld\n", (long long)1, n / 2, n / 2);
			continue;
		}
		else
		{
			long long p = n / 2;
			if (p % 2 == 0)
			{
				printf("%lld %lld %lld\n", n / 2, n / 4, n / 4);
				continue;
			}
			else
			{
				printf("%lld %lld %lld\n", (long long)2, (p - 1), (p - 1));
				continue;
			}
		}
	}
	return 0;
}

