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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

/**
 * Created by Валерий on 29.07.2016.
 */
public class ProblemB {
    public static void main(String[] args) throws IOException {
        long answer = 0;
        Map<Long, Long> map = new HashMap<>();
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(in.readLine());
        int n = Integer.parseInt(st.nextToken());
        long[] list = new long[n];
        st = new StringTokenizer(in.readLine());
        for (int i = 0; i < n; i++) {
            long current = Long.parseLong(st.nextToken());
            list[i] = current;
            if (map.get(current) == null) {
                map.put(current, (long) 1);
            } else {
                map.put(current, map.get(current) + 1);
            }
        }
        for (long i: list) {
            for (long j = 1; j < Math.pow(2, 32); j*= 2) {
                if (map.get(j - i) != null) {
                    answer += map.get(j - i);
                    if (j - i == i) {
                        answer--;
                    }
                }
            }
        }
        System.out.println(answer / 2);
    }
}
