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:
# cook your dish here
# cook your dish here
import math,sys
import heapq
from fractions import Fraction
from collections import Counter,defaultdict
def li(): return list(map(int,sys.stdin.readline().split()))
def ls(): return list(map(int,list(input())))
def la(): return list(input())
def ii():  return int(input())
def dic(x): return defaultdict(lambda: x) 
def isPrime(n):
    i= 2
    if n == 1:
        return False
    while i <= int(math.sqrt(n)):
        if n%i == 0:
            return False
        i = i + 1
    return True
def gcd(a,b):
    if b == 0:
        return a
    return gcd(b,a%b)
def LCM(a,b): 
    return (a*b) / gcd(a,b) 
t = int(input())
for _ in range(t):
    n,k = li()
    if (n- (k-1)*2)% 2  == 0 and (n- (k-1)*2) > 0:
        print('YES')
        a = '2 '*(k-1)+str((n- (k-1)*2))
        print(a)
    elif (n- (k-1))% 2  != 0 and (n- (k-1)) >0:
        print('YES')
        a = '1 '*(k-1)+str((n- (k-1)))
        print(a)
    else:
        print('NO')