CodeForces 702B Powers of Two [two points/lower_bound to find how many numbers/given an array to find the number pairs of ai + aj equal to the power of 2]

B. Powers of Two

 

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 
[Analysis]: After reading the question carefully, it is found that ai will not exceed 10^9, which is approximately equal to 2^30, so we can first type out a table of 2^k, and then for ai, subtract ai from 2^k, Use binary search to get the number of aj, and accumulate the result. The time complexity is O(N). 
【Code】:

#include<bits/stdc++.h>

using namespace std;
#define ll long long
#define maxn 100010
ll a[maxn];
int n;
/*
Given an array, find the number of pairs where ai + aj is a power of 2
*/ 
int main()
{

    while(cin >> n){
            ll ans = 0;
       for(int i=0; i<n; i++)
        cin >> a[i];
       sort(a,a+n);
       for(ll i=0; i<n; i++){
            for(ll j=0; j<32; j++){
                ll t = (1<<j) - a[i];
                ans += upper_bound(a+i+1,a+n,t)-lower_bound(a+i+1,a+n,t);
            }
       }
        cout << ans << endl;
    }
    return 0;
}

two points