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.*;

/**
 * Created by disqustingman on 13.08.16.
 */
public class PowerOfTwo {
    private static void initPowerOfTwo(List<Long> powerOfTwo){
        long curr = 1;
        while (curr < 2e9){
            powerOfTwo.add(curr);
            curr = curr * 2;
        }
    }
    public static void main(String[] args){
        Scanner inp = new Scanner(System.in);
        int n = inp.nextInt();
        Map<Long,Integer> numbers = new HashMap<>();
        List<Long> powersOfTwo = new ArrayList<>();
        long countPowerOfTwo = 0;
        initPowerOfTwo(powersOfTwo);

        for (int i = 0; i < n; i++){
            long val = inp.nextInt();
            for (Long powerOfTwo: powersOfTwo){
                long dif = powerOfTwo - val;
                if (numbers.containsKey(dif)){
                    countPowerOfTwo += numbers.get(dif);
                }
            }
            if (numbers.containsKey(val)){
                int count = numbers.get(val) + 1;
                numbers.put(val, count);
            }else{
                numbers.put(val, 1);
            }
        }

        System.out.println(countPowerOfTwo);
    }
}
