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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void write(LL a)
{
    if (a<0){printf("-");a=-a;}
    if (a>=10)write(a/10);
    putchar(a%10+'0');
}
inline void writeln(LL a){write(a);printf("\n");}
int n,sav;
LL ans;
int a[100010];
int rep[100010];
int bin[110];
inline bool bsearch(int l,int r,int x,int dat)
{
    if (dat<=0)return 0;
    if (l>r)return 0;
    int ans=-1;
    while (l<=r)
    {
        int mid=(l+r)>>1;
        if (a[mid]==dat){ans=mid;break;}
        if (a[mid]>dat)r=mid-1;
        if (a[mid]<dat)l=mid+1;
    }
    sav=ans;
    return (ans!=x||ans==x&&rep[x]>1)&&a[ans]==dat;
}
int main()
{
    n=read();
    bin[0]=1;
    for (int i=1;i<31;i++)bin[i]=bin[i-1]*2;
    for(int i=1;i<=n;i++)a[i]=read();
    sort(a+1,a+n+1);
    int cur=0;
    for(int i=1;i<=n;i++)
    {
        if (a[i]!=a[i-1])a[++cur]=a[i],rep[cur]=1;
        else rep[cur]++;
    }
    n=cur;
    for(int i=1;i<=n;i++)
    {
        for (int j=0;j<31;j++)
            if (bsearch(i,n,i,bin[j]-a[i]))
            {
                if (sav==i)ans+=(LL)rep[i]*(rep[i]-1)/2;
                else ans+=(LL)rep[i]*rep[sav];
            }
    }
    printf("%lld\n",ans);
}