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:
# link: https://codeforces.com/problemset/problem/1352/B

def check_odd(n,k):
    initial_sum = 1 * k
    last_number = 1 + (n - initial_sum)
    if last_number % 2 != 0 and last_number>0:
        return True
    else:
        return False     

def check_even(n,k):
    initial_sum = 2 * k
    last_number = 2 + (n - initial_sum)
    if last_number % 2 == 0 and last_number>0:
        return True
    else:
        return False    

if __name__ == "__main__":
    for _ in range(int(input())):
        n,k = map(int,input().split())
        if (n%2!=0 and k%2==0) or (k>n):
            print("no")
            continue
        if n%k == 0:
            print("yes")
            s = str(n//k)
            print("{} ".format(s)*k)
            continue
        if check_even(n,k):
            print("yes") 
            print("2 "*(k-1),end="")
            last_number = 2 + (n - (2*k))
            print(last_number)
        elif check_odd(n,k):
            print("yes")
            print("1 "*(k-1),end="")
            last_number = 1 + (n - (1*k))
            print(last_number)
        else:
            print("no")