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 PYTHON):

Solution:
#!/usr/bin/env python
"""
This file is part of https://github.com/Cheran-Senthil/PyRival.

Copyright 2018 Cheran Senthilkumar all rights reserved,
Cheran Senthilkumar <hello@cheran.io>
Permission to use, modify, and distribute this software is given under the
terms of the MIT License.

"""
from __future__ import division, print_function

import cmath
import itertools
import math
import operator as op
# import random
import sys
from atexit import register
from bisect import bisect_left, bisect_right
# from collections import Counter, MutableSequence, defaultdict, deque
# from copy import deepcopy
# from decimal import Decimal
# from difflib import SequenceMatcher
# from fractions import Fraction
# from heapq import heappop, heappush

if sys.version_info[0] < 3:
    # from cPickle import dumps
    from io import BytesIO as stream
    # from Queue import PriorityQueue, Queue
else:
    # from functools import reduce
    from io import StringIO as stream
    from math import gcd
    # from pickle import dumps
    # from queue import PriorityQueue, Queue


if sys.version_info[0] < 3:
    class dict(dict):
        """dict() -> new empty dictionary"""
        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)

        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)

        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)

    def gcd(x, y):
        """greatest common divisor of x and y"""
        while y:
            x, y = y, x % y
        return x

    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip


def sync_with_stdio(sync=True):
    """Set whether the standard Python streams are allowed to buffer their I/O.

    Args:
        sync (bool, optional): The new synchronization setting.

    """
    global input, flush

    if sync:
        flush = sys.stdout.flush
    else:
        sys.stdin = stream(sys.stdin.read())
        input = lambda: sys.stdin.readline().rstrip('\r\n')

        sys.stdout = stream()
        register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))


def main():
    n, k = map(int, input().split())
    bn = bin(n)[:1:-1]

    if (n < k) or (bn.count('1') > k):
        print('NO')
        exit()

    print('YES')

    r = k - bn.count('1')

    for i, j in enumerate(bn):
        if j != '1':
            continue

        if r == 0:
            print(2**i)
        else:
            if (2 ** i) <= r:
                print(*([1] * (2 ** i)))
                r -= (2 ** i) - 1
            else:
                p = (r + 1).bit_length() - 1
                print(*([2 ** (i - p)] * ((2 ** p) - ((r + 1) - (2 ** p)))))
                print(*([2 ** (i - p - 1)] * 2 * ((r + 1) - (2 ** p))))
                r = 0


if __name__ == '__main__':
    sync_with_stdio(False)
    main()