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;

/**
 *
 * @author Admin
 */
public class PowerOfIndex {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int b = in.nextInt();
            a[i] = b;
            Integer times = map.get(b);
            if (times == null) {
                map.put(b, 1);
            } else {
                map.put(b, times + 1);
            }
        }
        long count = calculate(map, a);
        System.out.println(count);
    }

    public static long calculate(HashMap<Integer, Integer> m, int[] a) {
        long count = 0;
        int n = a.length;
        for (int j = 0; j < n; j++) {
            int c = a[j];
            m.put(a[j], m.get(a[j]) - 1);
            int q = (int) Math.pow(2, 30);
            for (int k = q; k >= c; k = k / 2) {
                if (m.get(k - c) != null) {
                    count = count + (int) m.get(k - c);
                }
            }
        }
        return count;
    }
}
