B. Powers of Two

time limit per test 3 seconds

memory limit per test 256 megabytes

input standard input

output standard output


You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i. e. some integer xexists so that ai + aj = 2x).

Input

The first line contains the single positive integer n (1 ≤ n ≤ 105) — the number of integers.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2.

Examples

input

4
7 3 2 1


output

2

input


3
1 1 1



output

3


Note

In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4).

In the second example all pairs of indexes (i, j) (where i < j) include in answer.


题解：感觉没什么好说的。。。

AC代码：


#include<bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<map>
#include<queue>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MP(x,y) make_pair(x,y)  
template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }  
template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }
int read()
{
  int v = 0, f = 1;
  char c =getchar();
  while( c < 48 || 57 < c ){
    if(c=='-') f = -1;
    c = getchar();
  }
  while(48 <= c && c <= 57) 
    v = v*10+c-48, c = getchar();
  return v*f;
}
map<int,int>m;
int n,a;
ll ans;
int main()
{  
  cin>>n;
  for(int i =0; i<n; i++)
  {
    cin>>a;
    for(int j=1; j<=31; j++)
    {
      ans+=m[(1LL<<j)-a];          
    }    
    m[a]++;
  }
  printf("%I64d\n",ans);
  return 0;
}