Summarize to the Power of Two（map+思维）

A sequence d).

For example, the following sequences are good:

a3),
[1,1,1,1023],
[7,39,89,25,89],
[].

Note that, by definition, an empty sequence (with a length of 0) is good.

For example, the following sequences are not good:

aj such that their sum is a power of two),
aj such that their sum is a power of two),
aj such that their sum is a power of two).

You are given a sequence a1,a2,…,an. What is the minimum number of elements you need to remove to make it good? You can delete an arbitrary set of elements.

Input

The first line contains the integer 1≤n≤120000) — the length of the given sequence.

The second line contains the sequence of integers 1≤ai≤109).

Output

Print the minimum number of elements needed to be removed from the given sequence in order to make it good. It is possible that you need to delete all n elements, make it empty, and thus get a good sequence.

Examples

input

6
4 7 1 5 4 9

output

input

5
1 2 3 4 5

output

Copy

input

1
16

output

input

4
1 1 1 1023

output

Note

In he first example, it is enough to delete one element [4,7,1,4,9], which is good.

题目意思：给你一个数列，然后让你删掉某个数，让每一个数都可以与另外一个数相加，之和等于2的^d次方，问最少删掉的个数

解题思路：先预处理出2的次幂，然后暴力枚举出每个数与每个2的次幂之差，去判断在map中是否有这样的差存在（注意相同的两个数，比如4 ，4 这个时候就可以特判一下，出现次数），若不存在那么就需要删除掉这个数。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<map>
 4 #include<algorithm>
 5 #define ll long long int
 6 using namespace std;
 7 ll d[32];
 8 ll a[1000010];
 9 map<ll,ll>mp;
10 void double_2()
11 {
12     int i;
13     d[0]=1;
14     for(i=1; i<=31; i++)
15     {
16         d[i]=d[i-1]*2;
17     }
18 }
19 int main()
20 {
21     int i,j,flag,n;
22     ll k;
23     double_2();///2的幂
24     while(scanf("%d",&n)!=EOF)
25     {
26         int count=0;
27         mp.clear();///map清零
28         for(i=0; i<n; i++)
29         {
30             scanf("%lld",&a[i]);
31             mp[a[i]]++;///记录出现的次数
32         }
33         for(i=0; i<n; i++)
34         {
35             flag=0;
36             for(j=0; j<=31; j++)
37             {
38                 if(a[i]>=d[j])
39                 {
40                     continue;
41                 }
42                 else
43                 {
44                     k=d[j]-a[i];
45                     if(mp[k])///map中存在
46                     {
47                         if(a[i]==k)
48                         {
49                             if(mp[k]>=2)///map中需要至少两个
50                             {
51                                 flag=1;
52                                 break;
53                             }
54                         }
55                         else
56                         {
57                             flag=1;
58                             break;
59                         }
60                     }
61                 }
62             }
63             if(!flag)
64             {
65                 count++;
66             }
67         }
68         printf("%d\n",count);
69     }
70     return 0;
71 }