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:
"""
    This template is made by Satwik_Tiwari.
    python programmers can use this template :)) .
"""

#===============================================================================================
#importing some useful libraries.

import sys
import bisect
import heapq
from math import *
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl #
from bisect import bisect_right as br
from bisect import bisect

#===============================================================================================
#some shortcuts

mod = pow(10, 9) + 7
def inp(): return sys.stdin.readline().strip() #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)


#===============================================================================================
# code here ;))
def dfs(v,visited,st,new,cnt):
    print(cnt,st)
    new.append(st)
    visited[st] = 1
    for i in v[st]:
        if(visited[i] == 0):
            dfs(v,visited,i,new,cnt)
        else:
            cnt[0] +=1

def solve():
    n,k = sep()
    arr = []
    if(n%2==0):
        if(k>(n//2)):
            if(k>n or (k%2==1)):
                print('NO')
            else:
                print('YES')
                for i in range(0,k-1):
                    arr.append(1)
                arr.append(n-(k-1))
                print(*arr)
        else:
            print('YES')
            for i in range(0,k-1):
                arr.append(2)
            arr.append(n-(2*(k-1)))
            print(*arr)
    else:
        if(k%2==0 or (k>n)):
            print('NO')
        else:
            print('YES')
            for i in range(0,k-1):
                arr.append(1)
            arr.append(n-(k-1))
            print(*arr)

testcase(int(inp()))