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.Scanner;
import java.util.HashMap;

public class Solution {
    public static void main(String[] args) {
        Scanner inputScn = new Scanner(System.in);
        int n = inputScn.nextInt();
        int[] listNum = new int[n];
        HashMap<Integer, Integer> occurrences = new HashMap<Integer, Integer>();
        int highestNum = 0;

        for (int i = 0; i < n; i++) {
            listNum[i] = inputScn.nextInt();
            int prevOcc = occurrences.getOrDefault(listNum[i], 0);
            occurrences.put(listNum[i], prevOcc + 1);
            highestNum = Math.max(highestNum, listNum[i]);
        }

        long ans = 0;
        for (int curr : occurrences.keySet()) {
            for (long pow2 = 1; pow2 <= (highestNum << 1); pow2 += pow2) {
//                System.out.println(pow2 + "," + highestNum);
                if (pow2 < 2 * curr) continue;
                long freq = (long) occurrences.get(curr);
                if (pow2 == 2 * curr) {
                    long temp = (freq * (freq - 1)) / 2;
                    ans += temp;
                } else {
                    long freq2 = (long) occurrences.getOrDefault((int)(pow2 - curr), 0);
                    ans += freq2 * freq;
                }
//                System.out.println(curr + "," + (pow2 - curr));
            }
        }

        System.out.println(ans);
    }
}
