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.io.IOException;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.util.TreeMap;
public class PowersOfTwo {
	public static void main(String[]args)throws IOException,NumberFormatException{
		BufferedReader bf =new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.parseInt(bf.readLine());
		String s=bf.readLine();
		String[]sa=s.split(" ");
		Long []a=new Long[n];
		TreeMap<Long, Integer> t=new TreeMap<Long, Integer>();
		long []powers=new long[32];
		powers[0]=1;
		for(int i=1;i<32;i++){
			powers[i]=powers[i-1]*2;
		}
		for(int i=0;i<n;i++){
			a[i]=Long.parseLong(sa[i]);
			if(t.containsKey(a[i]))
				t.put(a[i],t.get(a[i])+1);
			else
				t.put(a[i], 1);
		}
		long cnt=0;
		for(int i=0;i<n;i++){
			for(int j=0;j<32;j++){
				if(powers[j]>a[i]){
					long other=powers[j];
					other-=a[i];
					if(t.containsKey(other)){
						cnt+=t.get(other);
						if(other==a[i])
							cnt--;
					}
				}
			}
		}
		System.out.println(cnt/2);
	}
}
