A sequence $a_1, a_2, \dots, a_n$ is called good if, for each element $a_i$, there exists an element $a_j$ ($i \ne j$) such that $a_i+a_j$ is a power of two (that is, $2^d$ for some non-negative integer $d$).

For example, the following sequences are good:
 -  $[5, 3, 11]$ (for example, for $a_1=5$ we can choose $a_2=3$. Note that their sum is a power of two. Similarly, such an element can be found for $a_2$ and $a_3$), -  $[1, 1, 1, 1023]$, -  $[7, 39, 89, 25, 89]$, -  $[]$. 
Note that, by definition, an empty sequence (with a length of $0$) is good.

For example, the following sequences are not good:
 -  $[16]$ (for $a_1=16$, it is impossible to find another element $a_j$ such that their sum is a power of two), -  $[4, 16]$ (for $a_1=4$, it is impossible to find another element $a_j$ such that their sum is a power of two), -  $[1, 3, 2, 8, 8, 8]$ (for $a_3=2$, it is impossible to find another element $a_j$ such that their sum is a power of two). 
You are given a sequence $a_1, a_2, \dots, a_n$. What is the minimum number of elements you need to remove to make it good? You can delete an arbitrary set of elements.
The first line contains the integer $n$ ($1 \le n \le 120000$) — the length of the given sequence.

The second line contains the sequence of integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$).
6
4 7 1 5 4 9
1
5
1 2 3 4 5
2
1
16
1
4
1 1 1 1023
0
10
2 10 9 1 10 4 7 8 5 4
5
2
1 1
0
2
1 6
2
6
1 7 7 7 7 7
0
3
1 2 3
1
3
1 3 3
0
2
3 3
2
2
3 1
0
3
1 2 2
1
2
2 2
0
2
2 1
2
3
1 1 3
0
3
1 3 2
1
3
1 1 2
1
1
1
1
def f(x):
    i = 0
    while x >= 2 ** i:
        i += 1
    return i

n = int(input())
l = list(map(int, input().split()))

u = {}
q = len(l)
mx = 0
for i in range(len(l)):
    try:
        u[l[i]] += 1
    except KeyError:
        u.update({l[i]: 1})
    if l[i] > mx:
        mx = l[i]

c = {}
for k, v in u.items():
    x = k
    try:
        c[x] += 0
        continue
    except KeyError:
        cl = f(x)
        was = cl
        while 2 ** cl - x <= mx:
            y = 2 ** cl - x

            if x == y:
                if v == 1:
                    cl += 1
                    continue
                q -= v
                c.update({x: 1})
                break

            try:
                u[y] += 0
            except KeyError:
                cl += 1
                continue
            q -= v
            try:
                c[y] += 0
            except KeyError:
                q -= u[y]
            c.update({x: 1})
            c.update({y: 1})
            break
print(q)
# Sat Oct 02 2021 14:25:53 GMT+0300 (Москва, стандартное время)
