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 
#!/usr/bin/env python3
import sys, getpass
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline  # to read input quickly

# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy

M9 = 10**9 + 7  # 998244353
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)]  # hexagonal layout
MAXINT = sys.maxsize

# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False  # codechef does not allow getpass
def log(*args):
    if OFFLINE_TEST:
        print('\033[36m', *args, '\033[0m', file=sys.stderr)

def solve(*args):
    # screen input
    if OFFLINE_TEST:
        log("----- solving ------")
        log(*args)
        log("----- ------- ------")
    return solve_(*args)

def read_matrix(rows):
    return [list(map(int,input().split())) for _ in range(rows)]

def read_strings(rows):
    return [input().strip() for _ in range(rows)]

# ---------------------------- template ends here ----------------------------


def solve_(n,k):
    # your solution here
    tail = k-3
    n -= tail

    if n%3 == 0:
        return [n//3, n//3, n//3] + [1]*tail
    if n%2 == 1:
        return [n//2, n//2, 1] + [1]*tail
    if n%4 == 0:
        return [n//4, n//4, n//2] + [1]*tail

    factor = 1
    while n%2 == 0:
        factor *= 2
        n = n//2
    return [n//2*factor, n//2*factor, factor] + [1]*tail


# for case_num in [0]:  # no loop over test case
# for case_num in range(100):  # if the number of test cases is specified
for case_num in range(int(input())):

    # read line as an integer
    # k = int(input())

    # read line as a string
    # srr = input().strip()

    # read one line and parse each word as a string
    # lst = input().split()
    
    # read one line and parse each word as an integer
    n,k = list(map(int,input().split()))
    # lst = list(map(int,input().split()))

    # read multiple rows
    # mrr = read_matrix(k)  # and return as a list of list of int
    # arr = read_strings(k)  # and return as a list of str

    res = solve(n,k)  # include input here
    
    # print result
    # Google and Facebook - case number required
    # print("Case #{}: {}".format(case_num+1, res))

    # Other platforms - no case number required
    # print(res)
    # print(len(res))
    print(*res)  # print a list with elements
    # for r in res:  # print each list in a different line
        # print(res)
        # print(*res)