codeforces 864B

 Polycarp and Letters

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 and lowercase;
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 the maximum number of elements 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.


Example
Input
11
aaaaBaabAbA
Output
2
Input
12
zACaAbbaazzC
Output
3
Input
3
ABC
Output
0
Note
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.

题意：找小写字母的子串中最多的不重复的字母个数。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
map<char,int>m;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        m.clear();
        string s;
        cin>>s;
        int ans=0,anss=0;
        for(int i=0; i<s.size(); i++)
        {
            if(s[i]>='A'&&s[i]<='Z')
            {
                m.clear();
                anss=max(anss,ans);
                ans=0;
                continue;
            }
            else if(s[i]>='a'&&s[i]<='z')
            {
                if(m[s[i]]==0)
                {
                    ans++;
                    anss=max(anss,ans);
                    m[s[i]]=1;
                }
            }
        }
        cout<<anss<<endl;
    }
}