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.InputStreamReader;
import java.util.StringTokenizer;


public class Main {

    static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st;

    public static void main(String[] args) throws Exception {
        int t = Integer.valueOf(in.readLine());
        for (int i = 0; i < t; i++) {
            st = new StringTokenizer(in.readLine());
            long n, k;
            n = Long.valueOf(st.nextToken());
            k = Long.valueOf(st.nextToken());
            if (n < k) {
                System.out.println("NO");
            } else {
                if (n % k == 0) {
                    System.out.println("YES");
                    for (int j = 1; j <= k; j++) {
                        System.out.print((n / k) + " ");
                    }
                } else {
                    long div = n / k;
                    long rem = n % k;
                    if (((div + rem) % 2 != 0 && div % 2 != 0)) {
                        System.out.println("YES");
                        for (long j = 1; j < k; j++) {
                            System.out.print(div + " ");
                        }
                        System.out.print((div + rem) + " ");
                    } else if (((div + rem) % 2 == 0 && div % 2 == 0)) {
                        System.out.println("YES");
                        for (long j = 1; j < k; j++) {
                            System.out.print(div + " ");
                        }
                        System.out.print((div + rem) + " ");
                    } else {
                        if ((k - 1) % 2 != 0 || div == 1) {
                            System.out.print("NO");
                        } else {
                            System.out.println("YES");
                            for (int j = 1; j <= k / 2; j++) {
                                System.out.print((div - 1) + " " + (div + 1) + " ");
                            }
                            System.out.print((div + rem) + " ");
                        }
                    }
                }
                System.out.println("");
            }
        }
    }
}