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.io.PrintWriter;
import java.util.StringTokenizer;

public class B {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static PrintWriter out = new PrintWriter(System.out);

	private static final String NO = "NO";
	private static final String YES = "YES";

	static void solve() throws Exception {
		long t = nextLong();
		for (int i = 0; i < t; i++) {
			long n = nextLong(), k = nextLong();
			long last;
			if (k > n || (n % 2 == 1 && k % 2 == 0)) {
				out.println(NO);
			} else if (n == k) {
				out.println(YES);
				out.print("1");
				for (int j = 0; j < k - 1; j++) {
					out.print(" 1");
				}
				out.println();
			} else if (n % 2 == 0 && k % 2 == 1) {
				last = n + 2 - 2 * k;
				if (last > 0) {
					out.println(YES);
					out.print(last);
					for (int j = 0; j < k - 1; j++) {
						out.print(" 2");
					}
					out.println();
				} else {
					out.println(NO);
				}
			} else {
				last = n + 1 - 1 * k;
				if (last > 0) {
					out.println(YES);
					out.print(last);
					for (int j = 0; j < k - 1; j++) {
						out.print(" 1");
					}
					out.println();
				} else {
					out.println(NO);
				}
			}

		}
	}

	public static void main(String[] args) throws Exception {
		solve();
		out.close();
	}

	static String next() throws Exception {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}

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

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