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 Rust (1.15.1) (judge verdict: Wrong Answer)

// Try AtCoder
// author: Leonardone @ NEETSDKASU

fn main() {
	let mut stdin = String::new();
    std::io::Read::read_to_string(
    	&mut std::io::stdin(),
        &mut stdin).unwrap();
	let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
    	($t:ty) => (get().parse::<$t>().unwrap());
        () => (get!(i64));
    }
    
    let n = get!();
    let mut s = std::collections::HashMap::new();
    
    for _ in 0..n {
    	let x = get!();
        *s.entry(x).or_insert(0) += 1;
    }
    
    let mut ks = vec![];
    for (k, _) in s.iter() {
    	ks.push(*k);
    }
    
    let mut ans = 0_i64;
    
    for &k in ks.iter() {
    	let mut p = 1;
        while p < k {
        	p <<= 1;
        }
        if p == k {
        	ans += s[&k] / 2;
        } else {
        	let e = p - k;
            let a = s[&k];
            let c = s.entry(e).or_insert(0);
            if *c <= 0 {
            	continue;
            }
            let t = std::cmp::min(a, *c);
            ans += t;
            *c -= t;
        }
    }
    
    println!("{}", ans);
    
}