Summarize to the Power of Two（map+思惟）

A sequence a1,a2,…,ana1,a2,…,an is called good if, for each element aiai, there exists an element ajaj (i≠ji≠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 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 a1,a2,…,ana1,a2,…,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 nn (1≤n≤1200001≤n≤120000) — the length of the given sequence.

The second line contains the sequence of integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤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 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 a4=5. The remaining elements form the sequence [4,7,1,4,9][4,7,1,4,9], which is good.

 

題目意思：給你一個數列，而後讓你刪掉某個數，讓每個數均可以與另一個數相加，之和等於2的^d次方，問最少刪掉的個數

解題思路：先預處理出2的次冪，而後暴力枚舉出每一個數與每一個2的次冪之差，去判斷在map中是否有這樣的差存在（注意相同的兩個數，好比4 ，4   這個時候就能夠特判一下，出現次數），若不存在那麼就須要刪除掉這個數。

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<map>
 4 #include<algorithm>
 5 #define ll long long int
 6 using namespace std;  7 ll d[32];  8 ll a[1000010];  9 map<ll,ll>mp; 10 void double_2() 11 { 12     int i; 13     d[0]=1; 14     for(i=1; i<=31; i++) 15  { 16         d[i]=d[i-1]*2; 17  } 18 } 19 int main() 20 { 21     int i,j,flag,n; 22  ll k; 23     double_2();///2的冪
24     while(scanf("%d",&n)!=EOF) 25  { 26         int count=0; 27         mp.clear();///map清零
28         for(i=0; i<n; i++) 29  { 30             scanf("%lld",&a[i]); 31             mp[a[i]]++;///記錄出現的次數
32  } 33         for(i=0; i<n; i++) 34  { 35             flag=0; 36             for(j=0; j<=31; j++) 37  { 38                 if(a[i]>=d[j]) 39  { 40                     continue; 41  } 42                 else
43  { 44                     k=d[j]-a[i]; 45                     if(mp[k])///map中存在
46  { 47                         if(a[i]==k) 48  { 49                             if(mp[k]>=2)///map中須要至少兩個
50  { 51                                 flag=1; 52                                 break; 53  } 54  } 55                         else
56  { 57                             flag=1; 58                             break; 59  } 60  } 61  } 62  } 63             if(!flag) 64  { 65                 count++; 66  } 67  } 68         printf("%d\n",count); 69  } 70     return 0; 71 }