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.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

public class Boss{
    public static void main(final String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        ArrayList <Long> pow = new ArrayList<Long>();
        long x = 1;
        while (x < 1e10) {
            pow.add(x);
            x = x * 2;
        }
        final ArrayList<Long> a = new ArrayList<Long>();
        for (int i = 0; i < n; i++) {
            a.add(s.nextLong());
        }
        long ans=0;
        HashMap <Long,Integer> hm=new HashMap<Long,Integer>();
        for(int i=0;i<n;i++){
            for(int j=0;j<pow.size();j++){
                if (hm.containsKey(pow.get(j)-a.get(i))){
                    ans+=hm.get(pow.get(j)-a.get(i));
                }
            }
            if (hm.containsKey(a.get(i))){
                hm.put(a.get(i),hm.get(a.get(i))+1);
            }
            else{
                hm.put(a.get(i),1);
            }
        }
        System.out.println(ans);
    }
}