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:
//package com.company;
import java.awt.*;
import java.util.*;

public class Main{

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tt = sc.nextInt();
        for(int t = 0 ; t < tt ; t ++){
            long n = sc.nextLong();
            long k = sc.nextLong();
            long ke = n;
            long ko = n;
            ArrayList<Long> odd = new ArrayList<Long>();
            ArrayList<Long> even = new ArrayList<Long>();
            for(long i = 1 ; i <= k-1; i++){
                odd.add((long)1);
                ke -= 1;
                even.add((long)2);
                ko -= 2;
            }
            odd.add(ke);
            even.add(ko);
            if(odd.get((int)k-1) < 0 && even.get((int)k-1) < 0){
                System.out.println("NO");
            }else{
                if(even.get((int)k-1) % 2 ==0 && even.get((int)k-1) > 0){
                    System.out.println("YES");
                    for(long i : even){
                        System.out.print(i+" ");
                    }
                    System.out.println();
                }else if(odd.get((int)k-1) % 2 != 0 && odd.get((int)k-1) > 0){
                    System.out.println("YES");
                    for(long j : odd){
                        System.out.print(j+" ");
                    }
                    System.out.println();
                }else{
                    System.out.println("NO");
                }
            }
        }
    }
}