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 math

t = int(input())
num = [input().split() for x in range(t)]

for x in num:
    i = 0
    j = 0
    tr = 0
    n = int(x[0])
    k = int(x[1])
    if n < k:
        print("NO")
        continue
    q = math.floor(n / k)
    r = n % k
    z = [q] * k
    z[-1] = q + r
    if (q + r) % 2 == 0 and q % 2 == 0:
        tr += 1
    elif (q + r) % 2 != 0 and q % 2 != 0:
        tr += 1
    elif (k - 1) % 2 == 0:
        tr += 2
    if tr == 1:
        print("YES")
        print(*z, sep=" ")
    elif tr == 2:
        for s in range(0, len(z) - 1, 2):
            z[s] += 1
            z[s + 1] -= 1
        if z[1] == 0:
            print("NO")
            continue
        print("YES")
        print(*z, sep=" ")
    elif tr == 0:
        print("NO")