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:
x=int(input())
m=[]
for f in range(x):
    m.append(input().split(" "))
for f in m:
    a = int(f[0])
    b = int(f[1])

    if (a % b == 0):
        print("YES")
        for i in range(b):
            print(a // b, " ", end='')
        print()
    elif (a//b)>0 and (a // b) % 2 == ((a // b) + (a % b)) % 2:
        print("YES")
        for i in range(b):
            print((a // b) + (a % b if i == b - 1 else 0), " ", end='')
        print()
    elif (b - 1) % 2 == 0 and (a//b)>1:
        print("YES")
        c = 1
        for i in range(b):
            print((a // b) + (a % b if i == b - 1 else c), " ", end='')
            c = c * -1
        print()
    else:
        print("NO")