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.HashMap;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;

public class PowerOfTwo {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int a = 1;
		Set<Integer> powerOf2 = new TreeSet<>();
		while(a>0 && a<Integer.MAX_VALUE){
			powerOf2.add(a);
			a = a*2;

		}

		int n = s.nextInt();
		HashMap<Integer, Integer> map = new HashMap<>();
		int[] array = new int[n];
		for(int i=0;i<n;i++){
			array[i] = s.nextInt();
		}

		long count = 0;
		for(int i=0;i<n;i++){
			for(int num: powerOf2){
				if(map.containsKey(num-array[i])){
					count = count + map.get(num-array[i]);
				}
			}
			if(map.containsKey(array[i])){
				map.put(array[i], map.get(array[i])+1);
			} else {
				map.put(array[i], 1);
			}
		}
		System.out.println(count);
	}

}