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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
int R=1,C=1,H[2000010],N[2000010];
//評価関数（いまはMIN）
int hyouka(int a,int b){
  if(C<b)return 1;
  if(C<a||b==0)return 0;
  return N[H[a]]<N[H[b]]?1:0;
}
//挿入関数
void hin(int a){
  int i=C++;
  for(N[H[0]=R]=a;hyouka(0,i/2);i/=2)H[i]=H[i/2];
  H[i]=R++;
}
//取り出す関数
int hout(){
  int rt=H[1],i,j=2,k=H[--C];
  for(i=1;hyouka(i,C);i=j)H[i]=H[j=i*2+1-hyouka(i*2,i*2+1)];
  H[j/2]=k;
  return rt;
}
int lb(int l,int r,int n,int *d){
  int m=(l+r)/2;
  if(l==r)return m;
  return n<=d[m]?lb(l,m,n,d):lb(m+1,r,n,d);
}
int main(){
  int n,i,j,d[100010],ans=0,a,f[100010]={};
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&a);
    hin(a);
  }
  for(i=0;i<n;i++)d[i]=N[hout()];
  for(i=0;i<n;i++){
    if(f[i])continue;
    f[i]=1;
    //printf("%d:",d[i]);
    for(j=0;j<32;j++){
      a=lb(0,n,(1<<(31-j))-d[i],d);
      //printf("%d ",a);
      if(a<n&&f[a]==0&&d[i]+d[a]==(1<<(31-j))){
	ans++;
	f[a]=1;
	break;
      }
    }//printf("\n");
  }
  printf("%d\n",ans);
  return 0;
}