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 PYTHON):

Solution:
import sys

__author__ = 'ratmir'

from collections import deque

alphabet = "abcdefghijklmnopqrstuvwxyz"

def solve(n, a, graph):
    return 1

def execute():
    [t] = [int(x) for x in sys.stdin.readline().split()]
    results = []
    for ti in range(0, t):
        [n, k] = [int(x1) for x1 in sys.stdin.readline().split()]

        t1 = n-k+1
        if t1%2==1 and t1>0:
            res = "YES\n"
            for i in range(0, k-1):
                res += "1 "
            res +=str((n-k+1))
        else:
            t2 = n-2*(k-1)
            if t2%2==0 and t2>0:
                res = "YES\n"
                for i in range(0, k-1):
                    res += "2 "
                res +=str((n-2*(k-1)))
            else:
                res = "NO"

        results.append(res+"\n")
    print(''.join('{}'.format(k) for k in results))


execute()