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.HashMap;
import java.util.Scanner;

public class powersoftwonew {
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		HashMap<Integer, Integer> hm = new HashMap<Integer,Integer>();
		int n = s.nextInt();
	    int[] ar = new int[n];
		for(int i=0;i<n;i++){
			int a = s.nextInt();
			ar[i] =a;
			if(!hm.containsKey(a))
				hm.put(a, 1);
			else {
				hm.put(a,hm.get(a)+1);
			}
		}// 2 6 6 6 6 6
		long count =0;
		for(int i=0;i<ar.length;i++){
			int ai=ar[i];
			if(hm.get(ai) !=0){
				hm.put(ai, hm.get(ai)-1);
			for(int x=1;x<=32;x++){
				int aj = (int) (1<<x)-ai;// && 
				if(hm.containsKey(aj) && hm.get(aj) !=0)
				  //System.out.println(hm.get(aj));
					count += hm.get(aj);
					
				
			}
		}
	}
		
		System.out.println(count);
	}

}
//YALLLAAAAAA YA SO2AL ZEBALAAAAAAAAAAAAAAAAAAAAAAA
