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:
for i in range(int(input())):
    a=list(map(int,input().strip().split()))[:2]
    v=[]
    if a[0]<a[1]:
        print("NO")
    else:
        if a[0] % 2 == 0 and a[1] <= int(a[0] / 2):
            print("YES")
            while a[1]>1:
                v.append(str(2))
                a[1]-=1
                a[0]=a[0]-2
            v.append(str(a[0]))
        elif a[0] % 2 == 0 and a[1] > int(a[0] / 2):
            if a[1] % 2 == 0:
                print("YES")
                while a[1] != 1:
                    v.append(str(1))
                    a[1] -= 1
                    a[0] -= 1
                v.append(str(a[0]))
            else:
                print("NO", end=" ")
        elif a[0] % 2 != 0 and a[1] % 2 == 0:
            print("NO", end=" ")
        else:
            print("YES")
            while a[1] != 1:
                v.append(str(1))
                a[1] -= 1
                a[0] -= 1
            v.append(str(a[0]))
        print(" ".join(v))