Codeforces-916B：Jamie and Binary Sequence (changed after round)（思维）

B. Jamie and Binary Sequence (changed after round)

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 (a1, a2, ..., ak) with 

.
Give a value 

 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 ≤ 1018, 1 ≤ k ≤ 105) —
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 [ - 1018, 1018].

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:

23 + 23 + 22 + 21 + 20 = 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:



Powers of 2:

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

If x = 0, then 2x = 1.

If x < 0, then 

.

Lexicographical order:

Given two different sequences of the same length, (a1, a2, ...
, ak) and (b1, b2, ...
, bk), the first one is smaller than the second one for the lexicographical order, if and only if ai < bi,
for the first i where ai and bi differ.

思路：我们可以先将n转化为二进制数，然后把高位的1转化低位的2个1，来满足总共有k个1。因为要保证最高位越小越好，一开始就不断从高位向低位转化。若再转化，1的个数将大于k时，就从低位开始一个一个转化，使得字典序尽量大。

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e6;
const int MOD=1e9+7;
typedef __int64 ll;
ll p[100],f[MAX];//f[i]表示2^(-i);
int main()
{
ll n,k,res=1;
cin>>n>>k;
int len=0,last=0;
while(n)p[len++]=n%2,n/=2;
for(int i=0;i<len;i++)k-=p[i];
if(k<0){puts("No");return 0;}//1的个数一开始就大于k
for(int i=len-1;;i--)
{
if((i>=0&&p[i]==0)||(i<0&&f[-i]==0))continue;
if((i>=0&&p[i]>k)||(i<0&&f[-i]>k))break;//再转化，1的个数将大于k
if(i>0)
{
p[i-1]+=2*p[i];
k-=p[i];
p[i]=0;
}
if(i==0)
{
f[1]+=2*p[i];
k-=p[i];
p[i]=0;
last=1;
}
if(i<0)
{
f[-(i-1)]+=2*f[-i];
k-=f[-i];
f[-i]=0;
last=-(i-1);
}
}
while(k)
{
for(int i=last;i>0;i--)//从低位开始
{
if(f[i]==0)continue;
k--;//一个一个的转化
f[i]--
c314
;
f[i+1]+=2;
last=i+1;
goto nex;
}
if(p[0])
{
k--;
p[0]--;
f[1]+=2;
last=1;
continue;
}
for(int i=1;i<len;i++)
{
if(p[i]==0)continue;
k--;
p[i]--;
p[i-1]+=2;
goto nex;
}
nex:;
}
puts("Yes");
for(int i=len-1;i>=0;i--)while(p[i]>0&&p[i]--)printf("%d ",i);
for(int i=1;i<=last;i++)while(f[i]>0&&f[i]--)printf("%d ",-i);
return 0;
}