【 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 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 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！！

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

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