cf702B 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 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.

n个数字，问有多少对数字加起来刚好是2的k次方。

这还用说？枚举个k再枚举个a[i]然后看看有没有2^k-a[i]这个数就好了

这里我为了防被x没用hash用了二分

不过要考虑一个数字出现很多次的情况，或者你要找的刚好就是这个数的情况

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define pi 3.1415926535897932384626433832795028841971
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 inline void write(LL a)
 {
     if (a<){printf("-");a=-a;}
     if (a>=)write(a/);
     putchar(a%+'');
 }
 inline void writeln(LL a){write(a);printf("\n");}
 int n,sav;
 LL ans;
 int a[];
 int rep[];
 int bin[];
 inline bool bsearch(int l,int r,int x,int dat)
 {
     if (dat<=)return ;
     if (l>r)return ;
     int ans=-;
     while (l<=r)
     {
         int mid=(l+r)>>;
         if (a[mid]==dat){ans=mid;break;}
         if (a[mid]>dat)r=mid-;
         if (a[mid]<dat)l=mid+;
     }
     sav=ans;
     return (ans!=x||ans==x&&rep[x]>)&&a[ans]==dat;
 }
 int main()
 {
     n=read();
     bin[]=;
     for (int i=;i<;i++)bin[i]=bin[i-]*;
     for(int i=;i<=n;i++)a[i]=read();
     sort(a+,a+n+);
     int cur=;
     for(int i=;i<=n;i++)
     {
         if (a[i]!=a[i-])a[++cur]=a[i],rep[cur]=;
         else rep[cur]++;
     }
     n=cur;
     for(int i=;i<=n;i++)
     {
         for (int j=;j<;j++)
             if (bsearch(i,n,i,bin[j]-a[i]))
             {
                 if (sav==i)ans+=(LL)rep[i]*(rep[i]-)/;
                 else ans+=(LL)rep[i]*rep[sav];
             }
     }
     printf("%lld\n",ans);
 }

cf702B