codeforces807div2 E.Prairie Partition[二分]

E. Prairie Partition

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

It can be shown that any positive integer x can be uniquely represented as x = 1 + 2 + 4 + ... + 2k - 1 + r, where k and r are integers, k ≥ 0, 0 < r ≤ 2k. Let's call that representation prairie partition of x.

For example, the prairie partitions of 12, 17, 7 and 1 are:

12 = 1 + 2 + 4 + 5,

17 = 1 + 2 + 4 + 8 + 2,

7 = 1 + 2 + 4,

1 = 1.

Alice took a sequence of positive integers (possibly with repeating elements), replaced every element with the sequence of summands in its prairie partition, arranged the resulting numbers in non-decreasing order and gave them to Borys. Now Borys wonders how many elements Alice's original sequence could contain. Find all possible options!

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of numbers given from Alice to Borys.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1012; a1 ≤ a2 ≤ ... ≤ an) — the numbers given from Alice to Borys.

Output

Output, in increasing order, all possible values of m such that there exists a sequence of positive integers of length m such that if you replace every element with the summands in its prairie partition and arrange the resulting numbers in non-decreasing order, you will get the sequence given in the input.

If there are no such values of m, output a single integer -1.

Examples

input

81 1 2 2 3 4 5 8

output

2 

input

61 1 1 2 2 2

output

2 3 

input

51 2 4 4 4

output

-1

Note

In the first example, Alice could get the input sequence from [6, 20] as the original sequence.

In the second example, Alice's original sequence could be either [4, 5] or [3, 3, 3].

题意：

一个整数可以拆分成2的幂次加一个尽可能小的常数，现给出一堆数字拆分的结果，问原来可能是几个数字拆来的。

思路：

原来数字的个数一定会小于等于拆分后1的个数。

如果原来可以由n个数字拆分得到，则一定也可以通过m个数字拆分得到（直接凑1），其中 n<m<=1的个数

所以只需要找到最小能拆分成几个即可。   且范围在1 到 pownum[0]之间，可以直接二分个数。

首先预处理记录2的幂次的个数到pownum[i], 2的i次幂与i+1次幂之间有多少个数字记录到betnum[i]

对于每一种拆分个数，针对当前链路进行操作，发生断链则进行相应操作，详细代码注释。

#include<bits/stdc++.h>using namespace std;typedef long long LL;LL n;int pownum[2005];   //pownum[i]表示2的i次方有多少个int betnum[2005];   //betnum[i]表示大于2的i次幂小于2的i+1次幂的个数int check(LL x){    //x表示当前的可更新的链的个数，初始一定为x，即总个数    LL t = pownum[0] - x; // 多余的1的个数    for(int i = 1; i < 64; ++i)    {        if(x >= pownum[i])        {            t -= min(t, x-pownum[i]);            x = pownum[i];        }        else            t += pownum[i]-x;        t += betnum[i];    }    return t<=x;}int main(){    ios::sync_with_stdio(false);    cin >> n;    LL x;    for(int i = 0; i < n; ++i)    {        cin >> x;        bool flag = 1;        int id = 0;        while(x>1)        {            if(x&1) // x不是2的幂次                flag = 0;            id++;            x >>= 1;        }        if(flag)            pownum[id]++;        else            betnum[id]++;    }    LL ans = -1, l = 1, r = pownum[0];    //二分个数， 最大为2^0的个数    while(l <= r)    {        LL mid = (l+r) >> 1;        if(check(mid))        {            ans = mid;            r = mid - 1;        }        else            l = mid + 1;    }    if(ans == -1)    {        cout << ans << endl;    }    else for(int i = ans; i <= pownum[0]; ++i)    {        cout << i << " ";    }    return 0;}