Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn't written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.

Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.

The first line of the input contains number n (1?≤?n?≤?100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn't exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.

Print a single integer — the maximum possible total length of words in Andrew's article.

In the first sample the optimal way to choose words is {'abb', 'aaa', 'bbb'}.

In the second sample the word 'cdecdecdecdecdecde' consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {'a', 'a', 'aaaa'}.

分析：自己没写出来，因为感觉太“麻烦”，后来看了下别人的代码，，才发现这是道超级暴力的题目

解答：暴力枚举选择的两个字母，然后分别统计在选择该两个字母的情况下所能得到的长度

复杂度分析：26*26*100*10^3=10^8  因此可以暴力而不会超时，可以数据范围能暗示解题方法

#include 
     
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        #include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               using namespace std; #define MM(a) memset(a,0,sizeof(a)) typedef long long LL; typedef unsigned long long ULL; const int maxn = 1e3+5; const int mod = 1000000007; const double eps = 1e-10; const int INF = 0x3f3f3f3f; LL gcd(LL a, LL b) { if(b == 0) return a; return gcd(b, a%b); } char a[105][1005]; int main() { int n; while(~scanf("%d",&n)) { int maxn=0; for(int i=1;i<=n;i++) scanf("%s",a[i]); for(int i=0;i<=25;i++) for(int j=0;j<=25;j++) //暴力枚举选择的两个字母 { if(j==i) continue; int num=0; for(int k=1;k<=n;k++) { bool flag=true; for(int h=0;h
              
               maxn) maxn=num; } printf("%d\n",maxn); } return 0; }