# 题目
ZOJ is 10 years old! For celebrating, we are offering the easiest problem in the world to you.

Recently we received a long sequence. We can modify the sequence once by the following two steps.

Choose any element in the sequence, say x(satisfying x ≥ 3), and subtract 3 from x.<br>
Choose any element in the sequence, say x, and add 1 to x.<br>
Now, we want to know how many times at most the sequence can be modified.<br>

Input

The input contains multiple test cases. For each case, the first line contains an integer <var>n</var>(1 ≤ <var>n</var> ≤ 20000). The second line contains <var>n</var> integers describing the sequence. All the numbers in the sequence are non-negative and not greater than 1000000.

Output

Output number of times at most the sequence can be modified, one line per case.

Sample Input
* 1
* 10
* 2
* 10 11

Sample Output
* 4
* 10

# 参考答案
    #include<iostream>
    #include<cstdlib>
    #include<cstring>
    #include<cstdio>

    using namespace std;

    long long x,ans,sum,cn2,cn1;
    int n;
    int main(){
     while(scanf("%d",&n)==1)
     {
     ans=0;
     cn2=0;
     cn1=0;
     for(int i=1;i<=n;i++)
     {
     scanf("%I64d",&x);
     ans+=x/3;
     if(x%3==2) cn2++;
     if(x%3==1) cn1++;
     }
     sum=ans;
     if(ans==0) {printf("0\n");continue;}
     ans+=cn2;
     if(sum<=cn1) ans+=(sum-1);
     else 
     {
     ans+=cn1;
     ans+=(sum-cn1-1)/2;
     }
     printf("%lld\n",ans);
     }
     return 0;
    }