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 
cases = int(input())
while cases:
    cases -= 1
    a, b = map(int, input().split())
    ans = []

    while b > 3:
        ans.append(1)
        b -= 1
        a-= 1

    if b == 3:
        if a % 2 != 0:
            print(a // 2, a // 2, 1, *ans)
        elif a % 4 == 0:
            print(a // 4, a // 4, a // 2, *ans)
        else:
            print(a // 2 - 1, a // 2 - 1, 2, *ans)

    # else:
    #
    #     if b % 2 != 0 and a % 2 != 0:
    #         md1 = a // 2
    #         md2 = a // 2 + 1
    #         fill = b - 2
    #         md1 -= fill//2
    #         md2 -= (fill+1)//2
    #         ans = [md1, md2] + [1]* fill
    #     elif b % 2 == 0 and a % 2 == 0:
    #         md1 = a // 2
    #         md2 = a // 2
    #         fill = b - 2
    #         md1 -= fill//2
    #         md2 -= fill // 2
    #         ans = [md1, md2] + [1]* fill
    #     elif a % 2 == 0 and b % 2 != 0:
    #         md1 = a // 2
    #         md2 = a // 2
    #         fill = b - 2
    #         md1 -= (fill+1)//2
    #         md2 -= (fill+1)//2
    #         ans = [md1, md2] + [1]* (fill-1) + [2]
    #     else:
    #         md1 = a // 2
    #         md2 = a // 2 + 1
    #         fill = b - 2
    #         md1 -= fill//2
    #         md2 -= (fill+1)//2
    #         ans = [md1, md2] + [1]* (fill-1) + [2]
    #
    #     print(*ans)







