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 (3.4.3) (judge verdict: Runtime Error)

import bisect

N=int(input())
A=list(map(int,input().split()))
A.sort()
B=[]
for i in range(1,10):
    if 2**i>2*(10**9):
        break
    else:
        B.append(2**i)
B.append(2**(i+1))
ans=0
#print(B)
for i in range(N):
    X=bisect.bisect_left(B,A[N-1-i])
    if B[X]==A[N-1-i]:
        X+=1
    #print(X)
    if B[X]==2*A[N-1-i]:
        #print(B[X],A[N-i])
        A[N-1-i]+=0.3
        #print(A[bisect.bisect_left(A,B[X]-A[N-i]+0.3)],B[X]-A[N-i]+0.3)
        if A[bisect.bisect_left(A,B[X]-A[N-1-i]+0.3)]==B[X]-A[N-1-i]+0.3:
            ans+=1
            #print(B[X],i)
            #print(B[X])
            A[bisect.bisect_left(A,B[X]-A[N-1-i]-0.2)]-=0.1
    elif A[bisect.bisect_left(A,B[X]-A[N-1-i])]==B[X]-A[N-1-i]:
        ans+=1
        #print(B[X],i)
        A[bisect.bisect_left(A,B[X]-A[N-1-i])]-=0.1
        A[N-1-i]+=0.2
print(ans)
#print(A)
#print(B)