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.util.LinkedList;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int count = in.nextInt();
        for (int i = 0; i < count; i++) {
            int n = in.nextInt();
            int k = in.nextInt();
            LinkedList<String> results = solve(n, k);
            if (results != null) {
                System.out.println("YES");
                System.out.println(String.join(" ", results));
            } else {
                System.out.println("NO");
            }
        }
    }

    public static LinkedList<String> solve(int n, int k) {
        LinkedList<String> results = new LinkedList<>();

        final boolean nIsEven = n % 2 == 0;
        final boolean kIsEven = k % 2 == 0;
        if (nIsEven) {
            if (kIsEven) {
                if (n < k) {
                    return null;
                }
                while (k > 1) {
                    k--;
                    results.add("1");
                    n -= 1;
                }
                results.add(String.valueOf(n));
            } else {
                //k is odd - therefore we cannot afford odd results => should check multiplication by 2
                if (n < k * 2) {
                    return null;
                }
                while (k > 1) {
                    k--;
                    results.add("2");
                    n -= 2;
                }
                results.add(String.valueOf(n));
            }
        } else {
            if (kIsEven) {
                return null;
            } else {
                if (n < k) {
                    return null;
                }
                while (k > 1) {
                    k--;
                    results.add("1");
                    n -= 1;
                }
                results.add(String.valueOf(n));
            }
        }

        return results;
    }
}