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;
import java.util.StringTokenizer;

public class SameParitySummands {

    public static void main(String[] args) {
        FastReader fs = new FastReader();
        StringBuilder sb = new StringBuilder();
        int t = fs.nextInt();
        for(int tt = 0; tt < t; tt++) {
            int n = fs.nextInt();
            int k = fs.nextInt();
            int num1 = n - (k - 1);
            int num2 = n - (k - 1) * 2;
            if(num1 >= 1 && num1 % 2 != 0) {
                sb.append("YES\n");
                for(int i = 0; i < k - 1; i++) {
                    sb.append("1 ");
                }
                sb.append(num1);
                sb.append("\n");
            }else if(num2 % 2 == 0 && num2 > 1) {
                sb.append("YES\n");
                for(int i = 0; i < k - 1; i++) {
                    sb.append("2 ");
                }
                sb.append(num2);
                sb.append("\n");
            }else {
                sb.append("NO\n");
            }
        }
        System.out.println(sb);
    }

    public static long div(long a, long b) {
        return (a + b - 1) / b;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {
            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());
        }

        float nextFloat() {
            return Float.parseFloat(next());
        }
    }
}