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.
//package contest702;

import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] arr = new int[n];
		for(int i = 0; i < n; i++){
			arr[i] = in.nextInt();
		}
		
		HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
		
		int[] powers = new int[50];
		int temp = 1; int pc=0;
		powers[pc] = temp;
		while(temp < 1000000000){
			temp*=2;
			pc++;
			powers[pc] = temp;
//			System.out.println(temp);
		}
		
		long count = 0;
		for(int i = 0; i < n; i++){
			for(int j = 0; j <= pc; j++){
				int toFind = powers[j]-arr[i];
				if(map.containsKey(toFind)){
					count+=map.get(toFind);
				}
			}
			if(map.containsKey(arr[i])){
				map.put(arr[i], map.get(arr[i])+1);
			}else{
				map.put(arr[i], 1);
			}
//			System.out.println(map.toString());
		}
		System.out.println(count);
	}
}
