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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class SameParitySummands {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());

        for (int i=0; i < t; i++) {
            String[] n = br.readLine().split(" ");
            solve(n);
        }

    }
    public static void solve(String[] input) {
        int n = Integer.parseInt(input[0]);
        int k = Integer.parseInt(input[1]);

        if (n < k) {
            System.out.println("NO");
            return;
        }
        //case #1 if n == k
        if (n==k) {
            System.out.println("YES");
            for (int i=0; i < n;i++) {
                System.out.print(1 + " ");
            }
            System.out.println();
            return;
        }


        //case #2 if the number is even
        if (n % 2 == 0) {
            int multi = k * 2 - 2;
            int multi2 = k - 1;
            if (n - multi > 0 ) {
                System.out.println("YES");
                for (int i=0; i < k-1; i++) {
                    System.out.print(2 + " ");
                }
                System.out.print(n - multi);
                System.out.println();
            }
            else if (n - multi2 > 0 && (n-multi2) % 2 == 1) {
                System.out.println("YES");
                for (int i = 0; i < k - 1; i++) {
                    System.out.print(1 + " ");
                }
                System.out.print(n - multi2 + " ");
                System.out.println();
            }
            else {
                System.out.println("NO");
            }
            return;
        }
        //case #3 if the number is odd. only possibility of yes is n and k are both odd
        if (n % 2 == 1 && k % 2 == 1) {
            int multi = k-1;
            if (n - multi > 0) {
                System.out.println("YES");
                for (int i = 0; i < k - 1; i++) {
                    System.out.print(1 + " ");
                }
                System.out.print(n - k + 1 + " ");
                System.out.println();
            }
        }
        else {
            System.out.println("NO");
        }
        return;

    }
}