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

11aaaaBaabAbA

output

2

input

12zACaAbbaazzC

output

3

input

3ABC

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.

解题说明：题意是给定一个长度为n的字符串，求最大的位置集合，每个位置都是小写字母且各不相同，每两个位置之间没有大写字母。可以采用暴力枚举来判断。

#include<cstdio>#include<iostream>#include<string>#include <ctype.h>#include<cstring>using namespace std;int occ[256];int main() {int n,i;int best = 0;int cnt = 0;scanf("%d", &n);getchar();unsigned char c;for (i = 0; i < n; ++i) {scanf("%c", &c);if (isupper(c)) {memset(occ, 0, 256 * 4);if (cnt > best){best = cnt;}cnt = 0;}else {cnt += (1 - occ[c]);occ[c] = 1;}}printf("%d\n", best > cnt ? best : cnt);return 0;}