二进制 + 模拟

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, ..., a_k) 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

Copy

23 5

Output

Yes
3 3 2 1 0 

Input

Copy

13 2

Output

No

Input

Copy

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 2^x = 2·2·2·...·2 (x times).

If x = 0, then 2^x = 1.

If x < 0, then .

Lexicographical order:

Given two different sequences of the same length, (a₁, a₂, ... , a_k) and (b₁, b₂, ... , 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.

题目分析 ： 将一个 n 分成 m 个2的幂次方的数，问是否存在这样的分法，如果存在，输出 yes , 并输出次幂，且保证其中的最大值最小，且字典序最大，首先我们的想法就是先采用最贪的取法，判断是否能够用 m 个数凑出来 n ，不能的话就是 no， 如果能为了让最大值最小，我们每次就将序列中最大数分成两份，知道分出 m 份，但是这样的分法只能确保出序列中的最大值最小，但是字典序却并不是最大的，因为 比如你输入 1 7 会得到 -2 -3 -3..全是 -3， 但是显然这样的字典序不是最大的，因此，我们最后再加一步对序列的合并，这里要怎么操作？首先序列如果只有一个值，我们是不用合并的，因为最初我们分序列是采用最贪的分法，所以除了最大值和第二大值，其他的值有且只会出现一次，我们合并就将第二大值合并，再将序列中最小的值一分为二即可。

代码示例 ：

#define ll long long

ll n, k;

ll fun(ll x){
    ll res = 1;
    ll base = 2;
    
    while(x > 0) {
        if (x & 1) res *= base;
        base = base * base;
        x >>= 1; 
    }
    
    return res;
}

map<ll, ll>mp;

int main() {
    cin >> n >> k;
    ll N = floor(log(n)/log(2));
    //printf("%lld
", len);
    
    ll sum = 0;
    ll cnt = 0;
    for(ll i = N; i >= 0; i--){
        ll p = fun(i);
        if (sum + p < n) {
            sum += p;
            mp[i]++;
            cnt++;
        } 
        else if (sum + p == n) {mp[i]++; cnt++; break;} 
    }
    if (cnt > k) {printf("No
"); return 0;}
    else printf("Yes
"); 
    map<ll, ll>::iterator it, ip;
    while (cnt < k) {
        it = --mp.end(); 
        ll f = it->first;
        mp[f-1] += 2;
        mp[f]--;
        if (mp[f] == 0) mp.erase(f);
        cnt++;
        //printf("*** %lld %lld %lld
", f, cnt, n);
    }
    
    it = mp.end();
    it--;
    int maxx = it->first;
    it--;
    ip = mp.begin();
    if (mp.size() != 1 && (it->second > 2 || (it->second == 2 && ip->first != it->first))) {
        while(it->second >= 2) {
            mp[maxx]++;
            mp[it->first] -= 2;
            ip = mp.begin();
            mp[ip->first]--;
            mp[ip->first-1] += 2;
            if (mp[it->first] == 0) mp.erase(it->first);
            if (mp[ip->first] == 0) mp.erase(ip->first);
        }
    }
    
    it = --mp.end();
    for(ip = it; ip != mp.begin(); ip--){
        for(ll j = 1; j <= ip->second; j++) printf("%lld ", ip->first);
        //printf("---
");
    }
    it = mp.begin();
    ll fff = ip->second;
    for(ll i = 1; i <= ip->second; i++){
        printf("%lld%c", it->first, i==ip->second?'
':' '); 
    }
    return 0;
}