Taxi

After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of s_i friends (1 ≤ s_i ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of groups of schoolchildren. The second line contains a sequence of integers s₁, s₂, ..., s_n (1 ≤ s_i ≤ 4). The integers are separated by a space, s_i is the number of children in the i-th group.

Output

Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.

Examples

Input

Copy

5
1 2 4 3 3

Output

Copy

4

Input

Copy

8
2 3 4 4 2 1 3 1

Output

Copy

5

Note

In the first test we can sort the children into four cars like this:

• the third group (consisting of four children),
• the fourth group (consisting of three children),
• the fifth group (consisting of three children),
• the first and the second group (consisting of one and two children, correspondingly).

There are other ways to sort the groups into four cars.

组合数是1,4    2,2   3,1   2,1,1     1,1,1,1  

#include
#include
#include
#include
using namespace std;
int main()
{
   int n,x;
   scanf("%d",&n);
   int sum=0;
   int a[5]={0};
   for(int i=0;i=a[1])
   {
       sum+=a[3];
       a[3]=0;             //比如是3，3，1的情况，3与1一辆，3单独一辆//
       a[1]=0;               
   }
   else
   {
       sum+=a[3];         //比如3，1，1，1的情况，3与1一辆，剩下的1的次数=1的次数-3的次数=2，咱们后面再在后面讨论这
       a[1]=a[1]-a[3];     //个a[1],如果a[1]>=4,就增加a[1]/4次，<4则增加1次。

   }
   if(a[1]>0)
   {
        if(a[1]%4==0)       //如果还有剩下的1，判断能否构成1，1，1，1组合 坐一辆车
    sum+=a[1]/4;             //若a[1]能被4整除，次数=a[1]/4   否则是a[1]/4+1
        else
    sum+=a[1]/4+1;
   }
   printf("%d\n",sum+a[4]);
   return 0;
}