You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i. e. some integer x exists so that ai + aj = 2x).

Input

The first line contains the single positive integer n (1 ≤ n ≤ 105) — the number of integers.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2.

Examples

Input

4
7 3 2 1


Output

2


Input

3
1 1 1


Output

3

Note

In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4).

In the second example all pairs of indexes (i, j) (where i < j) include in answer.
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;


public class two {
	public static void main(String args[])
	{
		Scanner get=new Scanner(System.in);
		Set<Integer> s=new HashSet<Integer>();
		HashMap<Integer,ArrayList<Integer> > hm=new HashMap<Integer,ArrayList<Integer> >();
		int n=get.nextInt();
		int a[]=new int[n];
		for(int i=0;i<n;i++)
		{
			int temp=get.nextInt();
			a[i]=temp;
			if(hm.containsKey(temp))
			{
				hm.get(temp).add(i);
			}
			else
			{
				ArrayList<Integer> al=new ArrayList<Integer>();
				al.add(i);
				hm.put(temp,al);
			}
		}
		long ans=0;
		for(int i=0;i<n;i++)
		{
			for(int j=1;j<32;j++)
			{
				int m=(int) Math.pow(2,j);
				if(a[i]>m)
					continue;
				if(hm.containsKey(m-a[i]))
				{
					ans+=find(hm.get(m-a[i]),i+1);
					//System.out.println(ans);
				}
			}
		}
		System.out.println(ans);
	}
	public static int find(ArrayList<Integer> al,int i)
	{
		if(i>al.get(al.size()-1))
			return 0;
		int index=bslast(al,0,al.size()-1,i);
			return al.size()-index;
	
	}
	public static int bslast(ArrayList<Integer> a,int low,int high,int element)
	{
	   
	    while(low<high)
	    {
	        int mid=low+(high-low)/2;
	         if(a.get(mid)>=element)
	            high=mid;
	        else
	            low=mid+1;
	    }
	    return low;
	}
}
