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.ArrayList;
import java.util.HashMap;
import java.util.Scanner;

public class Q2 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		HashMap<Integer, ArrayList<Integer>> map = new HashMap();
		int[] arr = new int[n];
		int max1 = 0;
		int max2 = 0;
		for (int i = 0; i < n; i++) {
			int read = sc.nextInt();
			arr[i] = read;
			if (max1 < read) {
				max2 = max1;
				max1 = read;
			} else if (max2 < read) {
				max2 = read;
			}
			if (map.containsKey(read)) {
				map.get(read).add(i);
			} else {
				ArrayList<Integer> temp = new ArrayList();
				temp.add(i);
				map.put(read, temp);
			}
		}
		long pair = 0;
		int upper = max1 + max2;
		int p = 0;
		while (upper >> p != 0) {
			p++;
		}
		ArrayList<Integer> candidate = new ArrayList();
		upper = (int) Math.pow(2, p - 1);
		while (upper != 0) {
			candidate.add(upper);
			upper >>= 1;
		}
//		System.out.println(candidate);
		for (int i = 0; i < n; i++) {
			for (int can : candidate) {
				if (can <= arr[i]) break;
				int diff = can - arr[i];
				if (map.containsKey(diff)) {
//					System.out.println(arr[i] + ":" + diff);
					ArrayList<Integer> index = map.get(diff);
					pair += index.size();
					if (arr[i] == diff) pair--;
				}
			}
		}
		System.out.println(pair / 2);
	}
}
