Problem: A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....

You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.

Input

The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).

Output

If it is impossible to represent n as the sum of k powers of two, print NO.

Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.

Examples

Input


9 4


Output


YES
1 2 2 4 


Input


8 1


Output


YES
8 


Input


5 1


Output


NO


Input


3 7


Output


NO

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class ProblemC {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		sc.close();
		if(k > n) {
			System.out.println("NO");
			System.exit(0);
		}else if(k == n) {
			System.out.println("YES");
			for(int i = 0; i < n; i++) {
				if(i == n - 1) {
					System.out.println(1);
				}else {
					System.out.print(1 +" ");
				}
			}
			System.exit(0);
		}
		String s = Integer.toBinaryString(n);
		int l = s.length();
		int b = Integer.bitCount(n);
		if(b > k) {
			System.out.println("NO");
			System.exit(0);
		}else if(b == k) {
			int[] a = new int[k];
			int idx = 0;
			for(int i = 0; i < l; i++) {
				if(s.charAt(i) == '1') {
					a[idx] = (int)Math.pow(2, l - i - 1);
					idx++;
				}
			}
			System.out.println("YES");
			for(int i = 0; i < k; i++) {
				if(i == k - 1) {
					System.out.println(a[i]);
				}else {
					System.out.print(a[i] +" ");
				}
			}
			System.exit(0);
		}
		int[] a = new int[k];
		int idx = 0;
		for(int i = 0; i < l; i++) {
			if(s.charAt(i) == '1') {
				a[idx] = (int)Math.pow(2, l - i - 1);
				idx++;
			}
		}
		int r = idx;
		loop: for(int i = 0; i < k; i++) {
			while(a[i] > 1) {
				if(a[r] != 0) break loop;
				a[i] /= 2;
				a[r] = a[i];
				r++;
				if(r == k) break loop;
			}
		}
		System.out.println("YES");
		disp(a);
	}
	static void disp(int[] a) {
		for(int i = 0; i < a.length; i++) {
			if(i == a.length - 1) {
				System.out.println(a[i]);
			}else {
				System.out.print(a[i] + " ");
			}
		}
	}
}