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:
from collections import deque
t = int(input())
rez = []

for i in range(t):
    p = ""
    A = input().split()
    n = int(A[0])
    k = int(A[1])
    if k == 1:
        rez.append("YES")
        rez.append(n)
    elif (n%2 != 0 and k%2 == 0) or k == n-1  or k > n:
        rez.append("NO")
        continue
    else:
        ost = int(n%k)
        ch = int(n//k)
        if n == k:
            for j in range(k):
                p+= "1 "
        elif ost % 2 == 0:
            for j in range(k-1):
                p += str(ch)+ " "
            p += str(ch + ost)
        elif ch == 1:
            rez.append("NO")
            continue
        else:
            ch -= 1
            for j in range(k-1):
                p += str(ch)+ " "
            p += str(n - (k-1)*ch)
        rez.append("YES")
        rez.append(p)


for i in range(len(rez)):
    print(rez[i])