#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<vector>
#include<cmath>
#include<cctype>
using namespace std;

typedef long long ll;
const int Maxn=610000;

int n,a[Maxn];
map<int,int> ma;

int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {
		scanf("%d",&a[i]);ma[a[i]]++;
	}int ans=0;
	sort(a+1,a+n+1);
	for(int i=n;i>=1;i--) {
		if(ma[a[i]]==0) continue;
		ma[a[i]]--;
		int l=1;
		while(l<=a[i]) l<<=1;
		if(ma[l-a[i]]!=0) {
			ans++;
			ma[l-a[i]]--;
		}
	}
	printf("%d",ans);
	return 0;
}
### Prompt

Please formulate a cpp solution to the following problem:

Takahashi has N balls with positive integers written on them. The integer written on the i-th ball is A_i. He would like to form some number of pairs such that the sum of the integers written on each pair of balls is a power of 2. Note that a ball cannot belong to multiple pairs. Find the maximum possible number of pairs that can be formed.

Here, a positive integer is said to be a power of 2 when it can be written as 2^t using some non-negative integer t.

Constraints

* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N


Output

Print the maximum possible number of pairs such that the sum of the integers written on each pair of balls is a power of 2.

Examples

Input

3
1 2 3


Output

1


Input

5
3 11 14 5 13


Output

2

### Response

```cpp
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<vector>
#include<cmath>
#include<cctype>
using namespace std;

typedef long long ll;
const int Maxn=610000;

int n,a[Maxn];
map<int,int> ma;

int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {
		scanf("%d",&a[i]);ma[a[i]]++;
	}int ans=0;
	sort(a+1,a+n+1);
	for(int i=n;i>=1;i--) {
		if(ma[a[i]]==0) continue;
		ma[a[i]]--;
		int l=1;
		while(l<=a[i]) l<<=1;
		if(ma[l-a[i]]!=0) {
			ans++;
			ma[l-a[i]]--;
		}
	}
	printf("%d",ans);
	return 0;
}
```