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: Accepted)

#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let N: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let A: Vec<i64> = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect()
    };

    // 大きい値から順に貪欲にペアを作成する
    let B = {
        let mut B = A.clone();
        B.sort();
        B
    };
    let mut s = std::collections::HashMap::new();
    for &a in &A { *(s.entry(a).or_insert(0)) += 1; }
    for &b in B.iter().rev() {
        let k = format!("{:b}", b).len();
        let c = (1 << k) - b;
        if b == c && *s.get(&b).unwrap_or(&0) >= 2 { *(s.entry(b).or_insert(0)) -= 2; }
        if b != c && *s.get(&b).unwrap_or(&0) > 0 && *s.get(&c).unwrap_or(&0) > 0 {
            *(s.entry(b).or_insert(0)) -= 1;
            *(s.entry(c).or_insert(0)) -= 1;
        }
    }
    let ans = (N - s.values().sum::<usize>()) / 2;

    println!("{}", ans);
}