Polycarp and Letters（set首战！）

Description

Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string s consisting only of lowercase and uppercase Latin letters.

Let A be a set of positions in the string. Let's call it pretty if following conditions are met:

letters on positions from A in the string are all distinct（与众不同的） and lowercase（小写字母的）;
there are no uppercase letters in the string which are situated between positions from A (i.e. there is no such j that s[j] is an uppercase letter, and a₁ < j < a₂ for some a₁ and a₂ from A).

Write a program that will determine（确定，下决心） the maximum number of elements(元素) in a pretty set of positions.

Input

The first line contains a single integer n (1 ≤ n ≤ 200) — length of string s.

The second line contains a string s consisting of lowercase and uppercase Latin letters.

Output

Print maximum number of elements in pretty set of positions for string s.

Sample Input

Input

11
aaaaBaabAbA

Output

Input

12
zACaAbbaazzC

Output

Input

3
ABC

Output

Hint

In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position.

In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements.

In the third example the given string s does not contain any lowercase letters, so the answer is 0.

题目意思：求连续的小写子母中，种类最多的是多少种。

解题思路：利用set中存储的元素的唯一性去重，第一次使用set，还是对set中的删除不是很熟悉，使用set的迭代器

 1 #include<cstring>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<set>
 5 using namespace std;
 6 char s[210];
 7 int main()
 8 {
 9     int n,i,sum,sum1;
10     char a;
11     sum=0;
12     sum1=0;
13     scanf("%d",&n);
14     getchar();
15     scanf("%s",s);
16     set<char>st;
17     set<char>::iterator it;
18     for(i=0; i<n; i++)
19     {
20         if(s[i]>='A'&&s[i]<='Z')
21         {
22             sum1=st.size();
23             for(it=st.begin(); it!=st.end();)
24             {
25                 st.erase(it++);
26             }
27             if(sum1>sum)
28             {
29                 sum=sum1;
30             }
31 
32         }
33         else if(s[i]>='a'&&s[i]<='z')
34         {
35             st.insert(s[i]);
36         }
37     }
38     sum1=st.size();
39     if(sum1>sum)
40     {
41         sum=sum1;
42     }
43     printf("%d\n",sum);
44 }

Polycarp and Letters（set首战！）

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