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;

public class PowersOfTwo {
	static ArrayList<Long> key;
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		key = new ArrayList<>();
		make();
		int n = scan.nextInt();
		HashMap<Long, Integer> fre = new HashMap<>();
		HashSet<Long> h = new HashSet<>();
		for(int i = 0 ; i < n ; i++){
			long a = scan.nextLong();
			if(!fre.containsKey(a)) fre.put(a, 1);
			else fre.put(a, fre.get(a)+1);
			h.add(a);
		}
		long ans = 0, alike = 0;
		for(long i : fre.keySet()){
			for(int j = 0 ; j < key.size() ; j++){
				if(h.contains(key.get(j)-i)){
					if(key.get(j)-i == i) alike+=(fre.get(i)*(fre.get(i)-1L))/2L;
					else ans+=fre.get(i)*(fre.get((key.get(j)-i)));
				}
			}
		}
		System.out.println((ans/2)+alike);
	}
	static void make(){
		long curr = 1;
		while(curr < 2000000000){
			key.add(curr);
			curr*=2;
		}
		key.add(curr);
	}
}
