C. Summarize to the Power of Two

A sequence a1,a2,…,ana1,a2,…,an is called good if, for each element aiai, there exists an element ajaj (i≠ji≠j) such that ai+ajai+aj is a power of two (that is, 2d2d for some non-negative integer dd).

For example, the following sequences are good:

  • [5,3,11][5,3,11] (for example, for a1=5a1=5 we can choose a2=3a2=3. Note that their sum is a power of two. Similarly, such an element can be found for a2a2 and a3a3),
  • [1,1,1,1023][1,1,1,1023],
  • [7,39,89,25,89][7,39,89,25,89],
  • [][].

Note that, by definition, an empty sequence (with a length of 00) is good.

For example, the following sequences are not good:

  • [16][16] (for a1=16a1=16, it is impossible to find another element ajaj such that their sum is a power of two),
  • [4,16][4,16] (for a1=4a1=4, it is impossible to find another element ajaj such that their sum is a power of two),
  • [1,3,2,8,8,8][1,3,2,8,8,8] (for a3=2a3=2, it is impossible to find another element ajaj such that their sum is a power of two).

You are given a sequence a1,a2,…,ana1,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 nn (1≤n≤1200001≤n≤120000) — the length of the given sequence.

The second line contains the sequence of integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤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 nn elements, make it empty, and thus get a good sequence.

Examples

input

Copy

6
4 7 1 5 4 9

output

Copy

1

input

Copy

5
1 2 3 4 5

output

Copy

2

input

Copy

1
16

output

Copy

1

input

Copy

4
1 1 1 1023

output

Copy

0

Note

In the first example, it is enough to delete one element a4=5a4=5. The remaining elements form the sequence [4,7,1,4,9][4,7,1,4,9], which is good.

 

题意:求至少删掉所给集合中的几个元素,使得集合内的任意一个元素x均能在剩余元素中找到一个元素y,使x+y为2的幂次方

思路:可以枚举所有2的幂次方,对每一个元素判断

#include
#include
#include
#include
#include
using namespace std;
int a[120009];
mapmp;
int main()
{
    int n;
    cin>>n;
    for(int i=0;i>a[i];
        mp[a[i]]++;
    }
    int sum=0;
    for(int i=0;i1||(mp[x]==1&&x!=a[i]))
            {
                sum++;
                break;
            }
        }
    }
    cout<

 

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