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 ≤ 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.

Sample Input

Input

51 2 4 3 3

Output

4

Input

82 3 4 4 2 1 3 1

Output

5

Hint

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.

代码如下：

package com.acm.njupt;import java.util.Arrays;import java.util.Scanner;public class B1 {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);    int n = scanner.nextInt();int num[] = new int[n];int i ;int count = 0;//num_4,num_3,num_2,num_1分别用来统计人数为4,3,2,1的组的个数int num_4 = 0 ,num_3 = 0 ,num_2 = 0,num_1 = 0;//分别用来存储人数为4,3,2,1的组所需要用到的出租车的数量int a4 = 0 , a3 = 0 , a2 = 0 ,a1 = 0 ;for( i = 0 ; i < n ; ++i ){num[i] = scanner.nextInt();}Arrays.sort(num);for(i = 0 ; i < n ;++i){if(num[i] == 4){++num_4;}if(num[i] == 3){++num_3;}if(num[i] == 2){++num_2;}if(num[i] == 1){++num_1;}}count += num_4;        if(num_3 >num_1){        count += num_3;                if(num_2%2 == 0){        count +=(num_2/2);        }else{        count +=(num_2/2 +1);        }        }else{        count += num_3;                num_1 -=num_3;                if(num_2%2 ==0){        count+=(num_2/2);        }else{        count += ((num_2/2) + 1);        if(num_1 >= 2){        num_1 -=2;        }else{        num_1 = 0;        }        }                if(num_1%4 == 0){        count+=(num_1/4);        }else{        count +=((num_1/4) + 1);        }        }                        System.out.println(count);}}