Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2 3

Sample Output 1

1

We can form one pair whose sum of the written numbers is 4 by pairing the first and third balls.
Note that we cannot pair the second ball with itself.

Sample Input 2

5
3 11 14 5 13

Sample Output 2

2

Solution in Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var
	a:array[0..35] of int64;
	b:array[0..200010] of longint;
	f:array[0..200010] of boolean;
	ans,i,j,n,x,y,z,l,r,mid:longint;
function check(x:longint):longint;
begin
	if x<b[1] then exit(-1);
	if x>b[n] then exit(-1);
	l:=1; r:=n;
	while l<=r do
	begin
		mid:=(l+r) div 2;
		if x=b[mid] then exit(mid);
		if x<b[mid] then r:=mid
		else l:=mid+1;
	end;
	exit(-1);
end;
procedure qsort(l,r:longint);
var
	i,j,t,mid:longint;
begin
	i:=l; j:=r; mid:=b[(l+r) div 2];
	repeat
		while b[i]<mid do inc(i);
		while b[j]>mid do dec(j);
		if i<=j then
		begin
			t:=b[i]; b[i]:=b[j]; b[j]:=t;
			inc(i); dec(j);
		end;
	until i>j;
	if l<j then qsort(l,j);
	if i<r then qsort(i,r);
end;
begin
	readln(n);
	for i:=1 to n do read(b[i]);
	a[1]:=2; ans:=0;
	for i:=2 to 31 do a[i]:=a[i-1]*2;
	qsort(1,n);
	fillchar(f,sizeof(f),true);
	y:=1; z:=1;
	for i:=1 to n-1 do
	begin
		while a[y]<b[i]+b[n] do inc(y);
		while a[z]<b[i] do inc(z);
		if f[i] then
			for j:=y downto z do
			begin
				x:=check(a[j]-b[i]);
				if (x<>-1) and ((x<>i) or (x=i) and (a[x-1]=a[i]) or (a[x+1]=a[i])) then 
				begin
					inc(ans);
					f[i]:=false;
					f[x]:=false;
					break;
				end;
			end;
	end;
	writeln(ans);
end.