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 Python (3.8.2) (judge verdict: Wrong Answer)

import sys
import bisect
import collections

sys.setrecursionlimit(10 ** 8)
ini = lambda: int(sys.stdin.readline())
inm = lambda: map(int, sys.stdin.readline().split())
inl = lambda: list(inm())
ins = lambda: sys.stdin.readline().rstrip()
debug = lambda *a, **kw: print("\033[33m", *a, "\033[0m", **dict(file=sys.stderr, **kw))

N = ini()
A = inl()
A.sort()


def solve():
    ac = collections.Counter(A)
    num_pairs = 0
    for i in range(N - 1, -1, -1):
        x = A[i]
        if ac[x] == 0:
            continue
        p = 1
        while p < x:
            p <<= 1
        ac[x] -= 1
        j = bisect.bisect_left(A, p - x, 0, i)
        y = A[j]
        if x + y != p:
            continue
        if ac[y] <= 0:
            continue
        ac[y] -= 1
        num_pairs += 1
    return num_pairs


print(solve())