Jamie is preparing a Codeforces round. He has got an idea for a problem, but does not know how to solve it. Help him write a solution to the following problem:

Find k integers such that the sum of two to the power of each number equals to the number n and the largest integer in the answer is as small as possible. As there may be multiple answers, you are asked to output the lexicographically largest one. 

To be more clear, consider all integer sequence with length k (a_1, a_2, ..., a_{k}) with $\sum_{i = 1}^{k} 2^{a_{i}} = n$. Give a value $y = \operatorname{max}_{1 \leq i \leq k} a_{i}$ to each sequence. Among all sequence(s) that have the minimum y value, output the one that is the lexicographically largest.

For definitions of powers and lexicographical order see notes.


-----Input-----

The first line consists of two integers n and k (1 ≤ n ≤ 10^18, 1 ≤ k ≤ 10^5) — the required sum and the length of the sequence.


-----Output-----

Output "No" (without quotes) in a single line if there does not exist such sequence. Otherwise, output "Yes" (without quotes) in the first line, and k numbers separated by space in the second line — the required sequence.

It is guaranteed that the integers in the answer sequence fit the range [ - 10^18, 10^18].


-----Examples-----
Input
23 5

Output
Yes
3 3 2 1 0 

Input
13 2

Output
No

Input
1 2

Output
Yes
-1 -1 



-----Note-----

Sample 1:

2^3 + 2^3 + 2^2 + 2^1 + 2^0 = 8 + 8 + 4 + 2 + 1 = 23

Answers like (3, 3, 2, 0, 1) or (0, 1, 2, 3, 3) are not lexicographically largest.

Answers like (4, 1, 1, 1, 0) do not have the minimum y value.

Sample 2:

It can be shown there does not exist a sequence with length 2.

Sample 3:

$2^{-1} + 2^{-1} = \frac{1}{2} + \frac{1}{2} = 1$

Powers of 2:

If x > 0, then 2^{x} = 2·2·2·...·2 (x times).

If x = 0, then 2^{x} = 1.

If x < 0, then $2^{x} = \frac{1}{2^{-x}}$.

Lexicographical order:

Given two different sequences of the same length, (a_1, a_2, ... , a_{k}) and (b_1, b_2, ... , b_{k}), the first one is smaller than the second one for the lexicographical order, if and only if a_{i} < b_{i}, for the first i where a_{i} and b_{i} differ.

Difficulty: HARD

Solutions:
import sys, os

def main():
	(n, k) = list(map(int, sys.stdin.readline().split()))
	x = bin(n)[2:]
	if x.count('1') > k:
		print('No')
		return
	ans = [0] * 10 ** 5
	val = len(x) - 1
	idx = len(x) - 1
	cnt = 0
	for i in x:
		if int(i):
			ans[val] = 1
			cnt += 1
		val -= 1
	for i in range(k - cnt):
		ans[idx] -= 1
		ans[idx - 1] += 2
		if ans[idx] == 0:
			idx -= 1
	maxIdx = idx - 1
	minIdx = idx - 1
	nonZeroIdx = idx - 1
	while 1:
		if minIdx < 0 and ans[minIdx] == 0:
			minIdx += 1
			break
		if ans[minIdx]:
			nonZeroIdx = minIdx
		minIdx -= 1
	minIdx = nonZeroIdx
	while 1:
		if ans[maxIdx] > 2 or (ans[maxIdx] == 2 and maxIdx != minIdx):
			ans[minIdx] -= 1
			ans[minIdx - 1] += 2
			ans[maxIdx] -= 2
			ans[maxIdx + 1] += 1
			minIdx -= 1
		else:
			maxIdx -= 1
		if maxIdx <= minIdx:
			break
	print('Yes')
	x = []
	while 1:
		for i in range(ans[idx]):
			x.append(idx)
		idx -= 1
		if idx < 0 and ans[idx] == 0:
			break
	print(' '.join(map(str, x)))

def __starting_point():
	main()
__starting_point()


ii = lambda : int(input())
kk = lambda : map(int, input().split())
ll = lambda : list(kk())
from math import log
elems = [0] * 126
(n, k) = kk()
c = 0
for i in range(63):
	if n & 2 ** i:
		elems[i] = 1
		c += 1
if c > k:
	print('No')
	exit()
for i in range(63, -63, -1):
	if elems[i]:
		if elems[i] > k - c:
			break
		c += elems[i]
		elems[i - 1] += elems[i] * 2
		elems[i] = 0
prin = []
for i in range(63, -63, -1):
	prin.extend([i] * elems[i])
while len(prin) < k:
	prin[-1] -= 1
	prin.append(prin[-1])
print('Yes')
print(' '.join(map(str, prin)))


import math
(x, k) = list(map(int, input().split()))
kori = k
a = bin(x)
qtz = 0
s = []
for i in range(2, len(a)):
	if a[i] == '1':
		k -= 1
		s.append(1)
	else:
		qtz += 1
		s.append(0)
v = []
for i in range(len(s)):
	if s[i] != 0:
		v.append(len(s) - 1 - i)
if k < 0:
	print('No')
	return
else:
	tam = len(s)
	print('Yes')
	if k > 0:
		p = 0
		while 1:
			if tam - 1 <= p:
				s.append(0)
			if s[p] > k:
				break
			else:
				k -= s[p]
				s[p + 1] += s[p] * 2
				s[p] = 0
				p += 1
		if k > 0:
			j = len(s) - 1
			while k > 0:
				while s[j] == 0:
					j -= 1
				s[j] -= 1
				if j == len(s) - 1:
					s.append(2)
					j += 1
				else:
					s[j + 1] += 2
					j += 1
				k -= 1
		v = []
		for i in range(len(s)):
			for j in range(s[i]):
				v.append(tam - 1 - i)
		print(*v)
	else:
		v = []
		for i in range(len(s)):
			for j in range(s[i]):
				v.append(len(s) - 1 - i)
		print(*v)


from collections import defaultdict

def solve(n, k):
	as_bin = bin(n)[2:]
	cnt = defaultdict(int)
	cnt.update({i: 1 for (i, b) in enumerate(reversed(as_bin)) if b == '1'})
	curr_len = len(cnt)
	curr_pow = len(as_bin) - 1
	if curr_len > k:
		return None
	while True:
		new_len = curr_len + cnt[curr_pow]
		if new_len > k:
			break
		cnt[curr_pow - 1] += 2 * cnt[curr_pow]
		del cnt[curr_pow]
		curr_pow -= 1
		curr_len = new_len
	i = min(cnt.keys())
	while curr_len < k:
		cnt[i] -= 1
		cnt[i - 1] += 2
		curr_len += 1
		i -= 1
	ans = []
	for i in sorted(cnt.keys(), reverse=True):
		ans.extend([i] * cnt[i])
	return ans
(n, k) = [int(v) for v in input().split()]
ans = solve(n, k)
if ans is None:
	print('No')
else:
	print('Yes')
	print(' '.join((str(c) for c in ans)))


(h, k) = [int(n) for n in input().split(' ')]
level = 0
ret = []
while h > 0:
	if h % 2 == 0:
		h //= 2
		level += 1
	else:
		ret.append(level)
		h -= 1
if len(ret) > k:
	print('No')
else:
	print('Yes')
	cntnum = {}
	maxn = ret[0]
	minn = ret[0]
	total_len = len(ret)
	for i in ret:
		if i not in cntnum:
			cntnum[str(i)] = 0
		cntnum[str(i)] += 1
		if maxn < i:
			maxn = i
		if minn > i:
			minn = i
	while total_len <= k:
		if total_len + cntnum[str(maxn)] <= k:
			if str(maxn - 1) not in cntnum:
				cntnum[str(maxn - 1)] = 0
			cntnum[str(maxn - 1)] += 2 * cntnum[str(maxn)]
			total_len += cntnum[str(maxn)]
			cntnum[str(maxn)] = 0
			maxn -= 1
			minn = min(minn, maxn)
		else:
			break
	while total_len < k:
		cntnum[str(minn - 1)] = 2
		cntnum[str(minn)] -= 1
		minn -= 1
		total_len += 1
	ans = []
	for (num, v) in cntnum.items():
		for i in range(0, v):
			ans.append(int(num))
	ans.sort(reverse=True)
	print(' '.join([str(x) for x in ans]))


(n, k) = map(int, input().split())
s = list()
b = list()
base = 0
while n > 0:
	s.append(n % 2)
	b.append(base)
	n //= 2
	base += 1
s.reverse()
b.reverse()
t = sum(s)
if t > k:
	print('No')
	exit(0)
pos = 0
while t < k:
	if pos + 1 == len(s):
		s.append(0)
		b.append(b[-1] - 1)
	if t + s[pos] <= k:
		t += s[pos]
		s[pos + 1] += 2 * s[pos]
		s[pos] = 0
		pos += 1
	else:
		for i in range(len(s) - 1, -1, -1):
			if s[i] > 0:
				while t < k:
					if i + 1 == len(s):
						s.append(0)
						b.append(b[-1] - 1)
					s[i] -= 1
					s[i + 1] += 2
					t += 1
					i += 1
				break
res = list()
for i in range(len(s)):
	for j in range(s[i]):
		res.append(b[i])
print('Yes')
print(' '.join(map(str, res)))