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;
inline long long read() {
  long long a = 0;
  int f = 0;
  char p = getchar();
  while (!isdigit(p)) {
    f |= p == '-';
    p = getchar();
  }
  while (isdigit(p)) {
    a = (a << 3) + (a << 1) + (p ^ 48);
    p = getchar();
  }
  return f ? -a : a;
}
int n, m, cnt, sum, a[50], ans[200005];
int main() {
  n = read();
  m = read();
  while (n) {
    if (n & 1) ++a[cnt], ++sum;
    n >>= 1;
    ++cnt;
  }
  if (sum > m) {
    puts("NO");
    return 0;
  }
  for (int i = cnt; i; --i) {
    if (!a[i]) continue;
    if (a[i] >= m - sum) {
      a[i] -= m - sum;
      a[i - 1] += (m - sum) << 1;
      sum = m;
      break;
    }
    sum += a[i];
    a[i - 1] += a[i] << 1;
    a[i] = 0;
  }
  if (sum < m) {
    puts("NO");
    return 0;
  }
  puts("YES");
  for (int i = 0; i <= cnt; ++i)
    for (int j = 1; j <= a[i]; ++j) printf("%d ", 1 << i);
  return 0;
}