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 PYTHON3):

Solution:

from collections import deque


def main():
  n, k = map(int, input().strip().split(" "))
  if k > n:
    print("NO")
    return
  factors = find_factors(n)
  if k < len(factors):
    print("NO")
    return
  if k == len(factors):
    print("YES")
    print(" ".join([str(x[0]) for x in factors]))
    return
  k_factors = deque()
  split_remaining = k - len(factors)
  for factor in factors:
    val, exp = factor
    if split_remaining > 0:
      split_factors = deque([(val, exp)])
      while True:
        if split_remaining == 0:
          break
        gsf = split_factors.popleft() # greatest_split_factor
        if gsf[1] == 0:
          split_factors.appendleft(gsf)
          break
        split_factors.append((gsf[0]//2, gsf[1]-1))
        split_factors.append((gsf[0]//2, gsf[1]-1))
        split_remaining -= 1
      for f in split_factors:
        k_factors.append(f[0])
    else:
      k_factors.append(val)
  print("YES")
  print(" ".join([str(x) for x in k_factors]))

def find_factors(n):
  import math
  factors = deque()
  exp = int(math.log2(n))
  factor = int(2**exp)
  while factor > 0:
    if n - factor >= 0:
      factors.append((factor, exp))
      n -= factor
    factor //= 2
    exp -= 1
  return factors
main()