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:
#!/usr/bin/env python3
#%% for atcoder uniittest use
import sys
input= lambda: sys.stdin.readline().rstrip()
def pin(type=int):return map(type,input().split())
def tupin(t=int):return tuple(pin(t))
#%%code
def resolve():
    N=int(input())
    for n in range(N):
        n,k=pin()
        if k%2==0:
            if n%2!=0 or k>n :print("NO");continue
            ans=[1]*k#parity==odd
            ans[0]=1+n-k
        else:
            if n%2==0:#parity==even
                ans=[2]*k
                if sum(ans)>n:print("NO");continue
                ans[0]+=n-(k*2)
            else:#parity=odd
                ans=[1]*k#parity==odd
                if sum(ans)>n:print("NO");continue
                ans[0]=1+n-k
        print("YES")
        print(*ans)
#%%submit!
resolve()