Takahashi has N balls with positive integers written on them. The integer written on the i-th ball is A_i. He would like to form some number of pairs such that the sum of the integers written on each pair of balls is a power of 2. Note that a ball cannot belong to multiple pairs. Find the maximum possible number of pairs that can be formed.

Here, a positive integer is said to be a power of 2 when it can be written as 2^t using some non-negative integer t.

Constraints

* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N


Output

Print the maximum possible number of pairs such that the sum of the integers written on each pair of balls is a power of 2.

Examples

Input

3
1 2 3


Output

1


Input

5
3 11 14 5 13


Output

2
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = Integer.parseInt(sc.next());
        int[] A = new int[N];
        TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
        for (int i=0;i<N;i++) {
            A[i] = Integer.parseInt(sc.next());
            if (map.containsKey(A[i])) {
                map.put(A[i], map.get(A[i])+1);
            } else {
                map.put(A[i], 1);
            }
        }

        int ans = 0;
        while (!map.isEmpty()) {
            int kk = map.lastKey();
            if (map.get(map.lastKey())>1) {
                map.put(map.lastKey(), map.get(map.lastKey())-1);
            } else {
                map.remove(map.lastKey());
            }
            String ss = Integer.toBinaryString(kk);
            // System.out.println((long)Math.pow(2, ss.length())-kk);
            int val = (int)((long)Math.pow(2, ss.length())-kk);
            if (map.containsKey(val)) {
                if (map.get(val)>1) {
                    map.put(val, map.get(val)-1);
                } else {
                    map.remove(val);
                }
                ans++;
            }
        }
        // System.out.println(map);
        System.out.println(ans);
    }
}