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)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    }};

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused)]
macro_rules! debug {
    ($($format:tt)*) => (write!(std::io::stderr(), $($format)*).unwrap());
}
#[allow(unused)]
macro_rules! debugln {
    ($($format:tt)*) => (writeln!(std::io::stderr(), $($format)*).unwrap());
}

trait Bisect<T> {
    fn lower_bound(&self, val: &T) -> usize;
    fn upper_bound(&self, val: &T) -> usize;
}

impl<T: Ord> Bisect<T> for [T] {
    fn lower_bound(&self, val: &T) -> usize {
        let mut pass = self.len() + 1;
        let mut fail = 0;
        while pass - fail > 1 {
            let mid = (pass + fail) / 2;
            if &self[mid - 1] >= val {
                pass = mid;
            } else {
                fail = mid;
            }
        }
        pass - 1
    }
    fn upper_bound(&self, val: &T) -> usize {
        let mut pass = self.len() + 1;
        let mut fail = 0;
        while pass - fail > 1 {
            let mid = (pass + fail) / 2;
            if &self[mid - 1] > val {
                pass = mid;
            } else {
                fail = mid;
            }
        }
        pass - 1
    }
}

fn solve() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {
        ($($format:tt)*) => (write!(out,$($format)*).unwrap());
    }
    input! {
        n: usize,
        a: [i64; n],
    }
    let mut a = a;
    a.sort();
    for i in 0..n {
        a[i] *= 2;
    }
    let mut ans = 0;
    for i in (0..n).rev() {
        if a[i] % 2 == 1 {
            continue;
        }
        let mut two = 1;
        while two <= a[i] {
            two *= 2;
        }
        let opp = two - a[i];
        let idx = a.upper_bound(&opp);
        if idx > 0 && idx <= i && a[idx - 1] == opp {
            a[idx - 1] += 1;
            a[i] += 1;
            ans += 1;
        }
    }
    puts!("{}\n", ans);
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}