Sergei B., the young coach of Pokemons, has found the big house which consists of nflats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number n is only connected with the flat number n - 1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
Input
The first line contains the integer n (1 ≤ n ≤ 100 000) — the number of flats in the house.
The second line contains the row s with the length n, it consists of uppercase and lowercase letters of English alphabet, the i-th letter equals the type of Pokemon, which is in the flat number i.
Output
Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
Examples
Input
3
AaA
Output
2
Input
7
bcAAcbc
Output
3
Input
6
aaBCCe
Output
5
Note
In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.
题目大意：
给你一个字符串，长度为n，让你找出包含所有字母的最短的子序列，看样例应该很容易看明白。
分析：先计算出所有的字母的种类，开一个数组开存放不同的字母，在这里长度开58即可，大写字母加小写字母一共52个,在ASCII码表中A和z的距离为57，用0表示A,57表示z，算出所有字母的种类为d. 然后遍历查询，查询的字母的种类等于d时，再对开头重复的字母或序列进行删除就行了，直接看代码吧！
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
char c[100011];
int v[58]; //0表示A，57表示z
int main()
{
int n;
while (~scanf ("%d",&n))
{
scanf ("%s",c);
int d= 0;
memset(v,0,sizeof(0));
for (int i = 0 ; i < n ; i++){
if (v[c[i]-'A']==0)
{
d++;
v[c[i]-'A']++;
}
}
int i,j;
int ans = INF;
int ant = 0;
memset(v,0,sizeof(0));
int s = 0;
for (i = 0 ; i < n ; i++)
{
int id = c[i]-'A';
if (v[id]==0)
ant++;
v[id]++;
if (ant == d) //首部前移
{
while (ant == d)
{
ans = min (ans , i - s + 1);
int t =c[s]-'A';
if (v[t] == 1) //此次前移就退出了
ant--;
v[t]--;
s++;
}
}
}
printf ("%d\n",ans);
}
return 0;
}