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:
import sys,math,bisect
from collections import Counter,defaultdict as dd
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var))
def l(): return list(sp())
def strl(): return list(strsp())
def sp(): return map(int, data().split())
def strsp(): return map(str, data().split())

for _ in range(int(data())):
    n,k= sp()
    if (n-(2*(k-1)))%2 == 0 and n-(2*(k-1))>0:
        print("YES")
        a=[2]*(k-1)
        a.append(n-2*(k-1))
        print(*a)
    elif (n-(k-1))%2 != 0 and n-(k-1)>0:
        print("YES")
        a=[1]*(k-1)
        a.append(n-(k-1))
        print(*a)
    else:
        print("NO")