B. Polycarp and Letters

B. Polycarp and Letters

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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.

其实只要看懂题了,就挺简单的.但是貌似我一开始是想多的.

只要每次碰到大写字母判断下就行,别忘记最后再来个特判.

#include <bits/stdc++.h>
#define N 205
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
int n;
string s;
bool a[26];
int main(){
    cin>>n;
    cin>>s;
    int len =s.length();
    int ma=0;
    for(int i=0;i<len;i++){
        if(s[i]<='z'&&s[i]>='a')
            a[s[i]-'a']=1;
        else{

            int ans=0;
            for(int i=0;i<26;i++){
              ans+=a[i];
            }
            mem(a);
            ma=max(ma,ans);
        }
    }
    int p=0;
    for(int i=0;i<26;i++)
      p+=a[i];
      ma=max(ma,p);
    cout<<ma<<endl;
  return 0;
}