B - Taxi（贪心）

Problem 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_ifriends (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

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.

解题思路：简单贪心。先排列，可以按升序排，也可以按降序排，无论按哪种方式，思路都一样。这里采用降序排。我们希望的是有更多组的人数（不分离每组里的人数）刚好凑成可以容纳4个人的车辆，这样租的车辆将会更少。于是用两个指针指向一头一尾，头指向人数较多的一组，尾指向人数较少的一组，如果当前两者相加比4大，说明头的这一组可以单独租一辆车，即num++（车辆数加1），i++（头指针往后移一位），尾指针不用处理；如果当前两者相加和不大于4，即sum<=4，车辆数先加1，同时尾指针往前移一位，因为移后尾指针指向（原来一组的前一组）当前一组的人数可能还可以凑合成一辆车，于是贪心往前，直到大于4为止。这样当两指针相等时，说明已经实现了贪心策略，此时的num即为最少车辆数。

AC代码：

 #include<bits/stdc++.h>
 using namespace std;
 int n,num=,sum=,a[];//num为车辆数
 bool cmp(int a,int b){return a>b;}
 int main(){
     cin>>n;
     for(int i=;i<n;++i)cin>>a[i];
     sort(a,a+n,cmp);//降序排列
     for(int i=;i<n;++i){//这里注意小于n，即前面往后移，后面往前贪心。
         sum=a[i]+a[n-];
         if(sum<=){//如果sum不大于4，说明可能还有几个组的人可以容纳进一辆车
             num++;n--;//车数先加1，队伍减1，再来看看能否再容纳一个组的人数
             while(sum+a[n-]<=){sum+=a[n-];n--;}//如果还可以容纳，那就n--；
         }
         else num++;//如果sum比4大，说明a[i]这个组的人可以单独坐一辆车
     }
     cout<<num<<endl;
     return ;
 }