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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const int MAX = 100000;
const long long LINF = 1000000000000000000;
int arr[35];
int main() {
  int N, K;
  scanf("%d %d", &N, &K);
  int t = 1;
  int idx = 0;
  if (N == 1 && K == 1) {
    printf("YES\n1\n");
    return 0;
  }
  while (t <= N) {
    t *= 2;
    idx++;
  }
  t /= 2;
  idx--;
  int temp = idx;
  int sum = 0;
  while (N) {
    if (N >= t) {
      N -= t;
      arr[idx]++;
      sum++;
    }
    t /= 2;
    idx--;
  }
  bool flag = false;
  while (temp != 0) {
    while (arr[temp] && sum != K) {
      arr[temp]--;
      arr[temp - 1] += 2;
      sum++;
    }
    if (sum == K) {
      flag = true;
      break;
    }
    temp--;
  }
  if (flag) {
    puts("YES");
    for (int i = temp; i >= 0; i--) {
      for (int j = 0; j < arr[i]; j++) printf("%d ", (int)pow(2, i));
    }
    puts("");
  } else
    puts("NO");
  return 0;
}