Problem Description

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

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 nn (1≤n≤120000) — the length of the given sequence.

The second line contains the sequence of integers a1,a2,…,an(1≤ai≤10^9).

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

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

題意：給出 n 個(gè)數(shù)，對(duì)于每個(gè)數(shù) ai，要求 ai+aj 都為2的冪，求不符合條件的個(gè)數(shù)

思路：

要判斷是否和為 2 的冪，因此可以貪心的從 2^30 向下進(jìn)行枚舉，如果數(shù)組中的每一個(gè)數(shù) ai，在數(shù)組中存在 j-ai 的話，則證明符合條件，反之則不符合。

因?yàn)轭}目的數(shù)據(jù)范圍較大，因此用?map 存儲(chǔ)。

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<set> #include<map> #include<stack> #include<ctime> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 1000005 #define MOD 123 #define E 1e-6 using namespace std; map<long long,long long> mp; map<long long,bool>vis; long long a[N]; int main() {int n;cin>>n;for(int i=0;i<n;i++){cin>>a[i];mp[a[i]]++;}sort(a,a+n);int cnt=0;for(int i=0;i<n;i++){if(vis[a[i]])continue;bool flag=true;for(long long j=1<<30;j>=1;j>>=1){if(j<a[i])break;if(mp.count(j-a[i])){if(j-a[i]==a[i]&&mp[j-a[i]]==1)continue;vis[a[i]]=1;vis[j-a[i]]=1;flag=false;break;}}if(flag)cnt++;}cout<<cnt<<endl;return 0; }

?

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