Problem: You are given two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).

In other words, find a_1, a_2, …, a_k such that all a_i>0, n = a_1 + a_2 + … + a_k and either all a_i are even or all a_i are odd at the same time.

If such a representation does not exist, then report it.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100).

Output

For each test case print:

  * YES and the required values a_i, if the answer exists (if there are several answers, print any of them); 
  * NO if the answer does not exist. 



The letters in the words YES and NO can be printed in any case.

Example

Input


8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9


Output


YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120

Solution (in PYTHON3):

Solution:
from math import *
from bisect import bisect_left, bisect_right
from queue import Queue
from sys import stdin, stdout
from collections import Counter
input = lambda: stdin.readline().strip()
fast_print = stdout.write

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    if (n%2==0 and (2*k<=n or (k%2==0 and k<=n))) or (n%2==1 and k%2==1 and k<=n):
        fast_print('YES\n')
        if k%2==0 and k<=n:
            for i in range(k-1):
                fast_print('1 ')
            fast_print(str(n-k+1))
        else:
            for i in range(k-1):
                fast_print('2 ') if n%2==0 else fast_print('1 ')
            if n%2==0:
                fast_print(str(n-2*(k-1)))
            else:
                fast_print(str(n-k+1))
        fast_print('\n')
    else:
        fast_print('NO\n')