题目链接    CodeForces - 1005C

A sequence a1,a2,…,an is called good if, for each element ai, there exists an element aj (i≠j) such that ai+aj is a power of two (that is, 2d for some non-negative integer d).

For example, the following sequences are good:[5,3,11]

(for example, for a1=5 we can choose a2=3. Note that their sum is a power of two. Similarly, such an element can be found for a2 and 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:[16]

(for a1=16, it is impossible to find another element aj

  • such that their sum is a power of two),
  • [4,16]
  • (for a1=4, it is impossible to find another element aj
  • such that their sum is a power of two),
  • [1,3,2,8,8,8]
  • (for a3=2, it is impossible to find another element 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 n(1≤n≤120000) — the length of the given sequence.

    The second line contains the sequence of integers a1,a2,…,an(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

    1

    Input

    5
1 2 3 4 5

    Output

    2

    Input

    1
16

    Output

    1

    Input

    4
1 1 1 1023

    Output

    0

    Note

    In the first example, it is enough to delete one element a4=5

    . The remaining elements form the sequence [4,7,1,4,9], which is good.

 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include <cstring>
#include <map>
using namespace std;
typedef long long LL;
int main()
{
    map<LL,LL> m;
    LL n,a[200000];
    LL i,k,ans=0;
    cin>>n;
    for(i=0;i<n;i++)
    {
        cin>>a[i];
        m[a[i]]++;
    }
    for(i=0;i<n;i++)
    {
        LL t=1,g=0,j=0;

        for(j=0;j<31;j++)
        {
            if(m[t-a[i]]!=0)
            {
                if(t-a[i]==a[i])
                {
                    if(m[a[i]]>1)
                        g++;
                    else
                        ;
                }
                else
                    g++;

            }


            t*=2;
        }
        if(g==0)
            ans++;
    }
    cout<<ans<<endl;
    return 0;

}