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

2

Input

12
zACaAbbaazzC

Output

3

Input

3
ABC

Output

0

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，当时使用set的迭代器也就是一个个删除的

for(it=st.begin(); it!=st.end();)
     {
          st.erase(it++);
     }

实际上set中含有直接清空容器的函数clear().

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<set>
using namespace std;
char s[210];
int main()
{
    int n,i,sum,sum1;
    char a;
    sum=0;
    sum1=0;
    scanf("%d",&n);
    getchar();
    scanf("%s",s);
    set<char>st;
    set<char>::iterator it;
    for(i=0; i<n; i++)
    {
        if(s[i]>='A'&&s[i]<='Z')
        {
            sum1=st.size();
            /*for(it=st.begin(); it!=st.end();)
            {
                st.erase(it++);
            }*/
            st.clear();
            if(sum1>sum)
            {
                sum=sum1;
            }
        }
        else if(s[i]>='a'&&s[i]<='z')
        {
            st.insert(s[i]);
        }
    }
    sum1=st.size();
    if(sum1>sum)
    {
        sum=sum1;
    }
    printf("%d
",sum);
}