B. 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.

Examples

Input

11
aaaaBaabAbA

Output

Input

12
zACaAbbaazzC

Output

Input

3
ABC

Output

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.

 1 #include <bits/stdc++.h>
 2 #define N 205
 3 #define mem(a) memset(a,0,sizeof(a))
 4 using namespace std;
 5 int n;
 6 string s;
 7 bool a[26];
 8 int main(){
 9     cin>>n;
10     cin>>s;
11     int len =s.length();
12     int ma=0;
13     for(int i=0;i<len;i++){
14         if(s[i]<='z'&&s[i]>='a')
15             a[s[i]-'a']=1;
16         else{
17 
18             int ans=0;
19             for(int i=0;i<26;i++){
20               ans+=a[i];
21             }
22             mem(a);
23             ma=max(ma,ans);
24         }
25     }
26     int p=0;
27     for(int i=0;i<26;i++)
28       p+=a[i];
29       ma=max(ma,p);
30     cout<<ma<<endl;
31   return 0;
32 }