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 
import sys


def input():
    return sys.stdin.readline().rstrip()


def input():
    return sys.stdin.readline().rstrip()


def slv_case3(n):
    p = n//4
    if n % 4 == 0:
        return p, p, 2*p
    elif n % 4 == 1:
        return 2*p, 2*p, 1
    elif n % 4 == 2:
        return 2*p, 2*p, 2
    else:
        return 2*p + 1, 2*p + 1, 1
    return


def slv():
    n, k = map(int, input().split())
    if n % 2 == k % 2:
        ans = [1]*(k - 2) + ([(n - k + 2)//2]*2)
        print(*ans)
        return
    else:
        a, b, c = slv_case3(n - k + 3)
        ans = [1]*(k - 3) + [a, b, c]
        print(*ans)
        return


def main():
    t = int(input())
    for i in range(t):
        slv()
    return


main()
