Educational Codeforces Round 15 B

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 数的和为2^x

 题解：标记每个数  枚举2^x 与a[i]取差值  统计mp[2^x-a[i]]  注意ans/2 注意开LL

/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^    ^ ^
 O      O
******************************/
//#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
#include<queue>
#include<cmath>
#define ll __int64
#define PI acos(-1.0)
#define mod 1000000007
using namespace std;
int n;
int a[100005];
map<int,int>mp;
int main()
{
    scanf("%d",&n);
    mp.clear();
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&a[i]);
        mp[a[i]]++;
    }
    int exm=1;
    ll ans=0;
    for(int i=1; i<=30; i++)
    {
        exm=exm<<1;
        for(int j=1; j<=n; j++)
        {
            if(mp[exm-a[j]])
            {
                mp[a[j]]--;
                ans=ans+mp[exm-a[j]];
                mp[a[j]]++;
            }
        }
    }
    cout<<ans/2<<endl;
    return 0;
}