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.Scanner;

/**
 * Created by saurabhgangwar on 3/8/16.
 */
public class B {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int len = sc.nextInt();
        long arr[] = new long[len];
        long pow2[] = new long[32];
        HashMap<Long, Integer> hashMap = new HashMap<Long, Integer>();
        for (int i = 0; i < len; i++) {
            arr[i] = sc.nextInt();
            if (hashMap.containsKey(arr[i])) {
                Integer c = hashMap.get(arr[i]) + 1;
                hashMap.put(arr[i], c);

            } else
                hashMap.put(arr[i], 1);

        }
        pow2[0] = 1;
        for (int i = 1; i < 32; i++) {
            pow2[i] = 2 * pow2[i - 1];
        }
        long ans = 0;
        for (int i = 0; i < len; i++) {
            for (int j = 1; j < 32; j++) {
                if (hashMap.containsKey(pow2[j] - arr[i])) {
                    if (arr[i] == pow2[j] - arr[i]) {
                        ans = ans + hashMap.get(pow2[j] - arr[i])-1;
                    } else {
                        ans = ans + hashMap.get(pow2[j] - arr[i]);
                    }
                }
            }
        }
        System.out.println(ans/2);

    }
}
