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.*;
import java.util.*;
import java.lang.*;

public class c1 {
    public static void main(String[] args) {
        MyScanner scan = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));

        int t=scan.nextInt();
        while(t-->0) {
            int n=scan.nextInt();
            int k=scan.nextInt();

            if(n<k) out.println("NO");
            else if(k%2==0&&n%2!=0) out.println("NO");
            else {
                if(n-(2*(k-1))>=2&&(n-(2*(k-1)))%2==0) {
                    out.println("YES");
                    for (int c=0;c<k-1;c++) {
                        out.print(2+" ");
                    }
                    out.println(n-(2*(k-1)));
                } else if(n-(k-1)>=1&&(n-(k-1))%2!=0) {
                    out.println("YES");
                    for (int c=0;c<k-1;c++) {
                        out.print(1+" ");
                    }
                    out.println(n-(k-1));
                } else {
                    out.println("NO");
                }
            }
        }
        out.close();
    }

    //scanner
    public static PrintWriter out;

    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    //methods
    public static int gcd(int a, int b) {
        if (b == 0)
            return a;
        if (a == 0)
            return b;
        return (a > b) ? gcd(a % b, b) : gcd(a, b % a);
    }

    public static int lcm(int a, int b) {
        return a * b / gcd(a, b);
    }

    public static boolean isprime(long a) {
        if (a == 0 || a == 1) {
            return false;
        }
        if (a == 2) {
            return true;
        }
        for (int i = 2; i < Math.sqrt(a) + 1; i++) {
            if (a % i == 0) {
                return false;
            }
        }
        return true;
    }

}