CF 1132 B. Taxi

Description

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 ≤ 1e5) — the number of groups of schoolchildren. The second line contains a sequence of integers s 1, s 2, ..., 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.

Example

input

5
1 2 4 3 3

output

4

input

8
2 3 4 4 2 1 3 1

output

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.

Solution

  思路：

统计1,2,3,4的出现次数，对于4的情况较好处理直接+1即可。3的情况下可以和1组合，所以3和1看做一个整体但分开计算，因为1和3的个数不一样，对于每个3直接+1，而对于剩下的1则是和2组合，(nums[1]+nums[2]*2+3)/4 这个3则是为了将不足4的又大于0的情况进1.

#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
int main(){
    ll n,sum=0,cnt=0;
    cin>>n;
    ll nums[5]={0};
    while(n--){
        ll temp;
        cin>>temp;
        nums[temp]++;
    }

    sum += nums[4]+nums[3];
    nums[1] = max((ll)0,nums[1]-nums[3]);
    sum += (nums[1]+nums[2]*2+3)/4;
    cout<<sum;
    return 0;
}