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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class PowTwo {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer stringTokenizer = new StringTokenizer(bufferedReader.readLine());
        final int powRange = 31;
        int n = Integer.parseInt(stringTokenizer.nextToken());
        int k = Integer.parseInt(stringTokenizer.nextToken());

        int[] pow2 = new int[powRange];
        pow2[0] = 1;
        for (int i = 1; i < powRange; i++) {
            pow2[i] = 2 * pow2[i - 1];
        }
        int[] countPow = new int[powRange];
        int num = n;
        int curMaxPow = powRange - 1;
        int curK = 0;
        for (int i = curMaxPow; i >= 0 && num > 0; i--) {
            if (pow2[i] <= num) {
                num -= pow2[i];
                countPow[i]++;
                curK++;
            }
        }
        int curPow = 30;
        while (curK < k && curPow > 0) {
            if (countPow[curPow] > 0) {
                countPow[curPow]--;
                countPow[curPow - 1] += 2;
                curK++;
            }
            if (countPow[curPow] == 0) {
                curPow--;
            }
        }
        if (curK == k) {
            System.out.println("YES");
            for (int i = 0; i < 30; i++) {
                for (int j = 0; j < countPow[i]; j++) {
                    System.out.print(pow2[i] + " ");
                }
            }
        } else {
            System.out.println("NO");
        }
    }
}