Summarize to the Power of Two（map+思维）

A sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an is\; called\; good\; if,\; for\; each\; element aiai,\; there\; exists\; an\; element ajaj (i\ne ji\ne j)\; such\; that ai+ajai+aj is\; a\; power\; of\; two\; (that\; is, 2d2d for\; some\; non-negative\; integer dd).}}^{}$

For example, the following sequences are good:

• $[5,3,11][5,3,11] (for\; example,\; for a1=5a1=5 we\; can\; choose a2=3a2=3.\; Note\; that\; their\; sum\; is\; a\; power\; of\; two.\; Similarly,\; such\; an\; element\; can\; be\; found\; for a2a2 and a3a3),$
• $[1,1,1,1023][1,1,1,1023],$
• $[7,39,89,25,89][7,39,89,25,89],$
• $[][].$

Note that, by definition, an empty sequence (with a length of $\mathrm{00)\; is\; good.}$

For example, the following sequences are not good:

• $[16][16] (for a1=16a1=16,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two),$
• $[4,16][4,16] (for a1=4a1=4,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two),$
• $[1,3,2,8,8,8][1,3,2,8,8,8] (for a3=2a3=2,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two).$

You are given a sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an.\; What\; is\; the\; minimum\; number\; of\; elements\; you\; need\; to\; remove\; to\; make\; it\; good?\; You\; can\; delete\; an\; arbitrary\; set\; of\; elements.}}^{}$

Input

The first line contains the integer $\mathrm{nn (1\le n\le 1200001\le n\le 120000)\; —\; the\; length\; of\; the\; given\; sequence.}$

The second line contains the sequence of integers ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109).}}^{}$

Output

Print the minimum number of elements needed to be removed from the given sequence in order to make it good. It is possible that you need to delete all $\mathrm{nn elements,\; make\; it\; empty,\; and\; thus\; get\; a\; good\; sequence.}$

Examples

input

6
4 7 1 5 4 9

output

1

input

5
1 2 3 4 5

output

Copy

2

input

1
16

output

1

input

4
1 1 1 1023

output

0

Note

In he first example, it is enough to delete one element ${\mathrm{a4=5.\; The\; remaining\; elements\; form\; the\; sequence [4,7,1,4,9][4,7,1,4,9],\; which\; is\; good.}}^{}$

 

题目意思：给你一个数列，然后让你删掉某个数，让每一个数都可以与另外一个数相加，之和等于2的^d次方，问最少删掉的个数

解题思路：先预处理出2的次幂，然后暴力枚举出每个数与每个2的次幂之差，去判断在map中是否有这样的差存在（注意相同的两个数，比如4 ，4   这个时候就可以特判一下，出现次数），若不存在那么就需要删除掉这个数。

 

#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
#define ll long long int
using namespace std;
ll d[32];
ll a[1000010];
map<ll,ll>mp;
void double_2()
{
    int i;
    d[0]=1;
    for(i=1; i<=31; i++)
    {
        d[i]=d[i-1]*2;
    }
}
int main()
{
    int i,j,flag,n;
    ll k;
    double_2();///2的幂
    while(scanf("%d",&n)!=EOF)
    {
        int count=0;
        mp.clear();///map清零
        for(i=0; i<n; i++)
        {
            scanf("%lld",&a[i]);
            mp[a[i]]++;///记录出现的次数
        }
        for(i=0; i<n; i++)
        {
            flag=0;
            for(j=0; j<=31; j++)
            {
                if(a[i]>=d[j])
                {
                    continue;
                }
                else
                {
                    k=d[j]-a[i];
                    if(mp[k])///map中存在
                    {
                        if(a[i]==k)
                        {
                            if(mp[k]>=2)///map中需要至少两个
                            {
                                flag=1;
                                break;
                            }
                        }
                        else
                        {
                            flag=1;
                            break;
                        }
                    }
                }
            }
            if(!flag)
            {
                count++;
            }
        }
        printf("%d\n",count);
    }
    return 0;
}