Educational Codeforces Round 15_B. Powers of Two

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 x exists 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.

题意：

给你n个数，问你有多少对满足a[i]+a[j]为2的次方.

题解：

首先，我们考虑他们能加起来的sum最多也就32个，0~2^31，所以我们对每一个a[i]都枚举sum-a[i],然后二分找这个数集里面有多少个满足条件，这里要考虑特殊的情况，如果sum-a[i]=a[i],那么你找到满足条件的个数要减1，然后最后ans要除2，因为sum-a[i]=a[j],sum-a[j]=a[i]，算了2次

 #include<bits/stdc++.h>
 #define F(i,a,b) for(int i=a;i<=b;++i)
 using namespace std;
 typedef long long ll;

 const int N=1e5+;
 int a[N];
 int main(){
     int n;
     ll ans=;
     scanf("%d",&n);
     F(i,,n)scanf("%d",a+i);
     sort(a+,a++n);
     F(i,,n)
     {
         for(int j=;j>=;j--){
             int now=<<j;
             if(now<a[i])break;
             int tmp=now-a[i];
             int pos1=lower_bound(a+,a++n,tmp)-a;
             int pos2=upper_bound(a+,a++n,tmp)-a;
             if(a[pos1]==tmp){
                 ans+=pos2-pos1;
                 if(tmp==a[i])ans--;
             }
         }
     }
     ans/=;
     printf("%I64d\n",ans);
     return ;
 }