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 (1.42.0) (judge verdict: Wrong Answer)

use std::cmp::*;
use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let mut a: Vec<u64> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();
    a.sort_by(|a, b| b.cmp(a));
    // でかい方からやると
    // a_i + something <= a_i * 2
    // aよりでかい

    let mut map = std::collections::HashMap::<u64, i64>::new();
    for i in 0..n {
        let p = map.entry(a[i]).or_insert(0);
        *p += 1;
    }

    let mut ans = 0;
    for i in 0..n {
        let j = (a[i] + 1).next_power_of_two();
        let p = map.entry(j - a[i]).or_insert(0);
        if j - a[i] == a[i] {
            if *p > 1 {
                ans += 1;
                *p -= 2;
            }
        } else if *p > 0 {
            ans += 1;
            *p -= 1;
        }
    }
    println!("{}", ans);
}