【 CodeForces - 864B】Polycarp and Letters（水题，字符串，有坑）

题干：

Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string sconsisting 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 fromAin the string are all distinct and lowercase;
• there are no uppercase letters in the string which are situated between positions fromA(i.e. there is no suchjthats[j] is an uppercase letter, anda1 <j<a2 for somea1 anda2 fromA).

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 6and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1and 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.

题目大意：

   就是说给一个字符串问你有多少 在不被大写字母分割的子串中（也就是不含大写字母的子串中），小写字母的种类数最多  有多少个。

解题报告：

    反正数据量很小，，随便写就行。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int bk[MAX]; 
int cnt[505];
int n,tmp;
char s[505];
int main()
{
  cin>>n;
  cin>>s+1;
  int len = strlen(s+1),ans = 0,tmp=0;
  for(int i = 1; i<=len; i++) {
    if(isupper(s[i])) {
      memset(cnt,0,sizeof cnt);
      ans = max(ans,tmp);
      tmp=0;
      continue;
    }
    if(cnt[s[i]] == 0) {
      tmp++;
      cnt[s[i]]++;
    }
    
  }
  printf("%d\n",max(tmp,ans));
  return 0 ;
 }

总结：

   注意最后输出的时候也需要取max，因为万一没进那个if(cnt[s[i]] == 0)这个if的话，，比如输入：x    那就凉凉啊。所以小地方一定要注意，，，还有一种处理办法就是每个tmp++之后就ans更新一下。。。还是那句话 一定要小范围测试一下！！