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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class Codeforces {

	
	
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		int n = Integer.parseInt(br.readLine());
		long [] in = new long[n];
		StringTokenizer st = new StringTokenizer(br.readLine());
		for (int i = 0 ; i< n ; i++){
			in[i] = Long.parseLong(st.nextToken());
		}
		HashMap<Long,Long> occ = new HashMap<>(); 
		long ans = 0;
		for (int i = 0 ; i< n ; i++){
			long current = in[i];
			for (int j = 0 ; j<31 ; j++){
				long power = 1<<j;
				long need = (power-current);
				if (occ.containsKey(need)){
					ans += occ.get(need);
				}
			}
			if (!occ.containsKey(current)){
				occ.put(current, 1l*1);
			}
			else 
			{
				long temp = occ.remove(current);
				occ.put(current,temp+1);
			}
		}
		out.println(ans);
		
		
		
		
		out.flush();
		out.close();
	}
}