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.

題目大意：又n個房間，從左到右排列，每個房間有一隻口袋妖怪，相鄰的房間時相通的， 每個房間只能進一次，問捕獲所有種類的口袋妖怪進入房間的最小個數

解題思路：直接使用尺取法的模板就行了，注意r走到L最右邊但不是最優解，L還可以往右移動這種情況，所以我們可以使用兩重迴圈來解決

AC程式碼：

#include<iostream>
#include<map>
#include<set>
using namespace std;
const int maxn=1e5+10;
int main()
{
	set<char> def;
	map<char,int>num;
	int n;
	char a[maxn],ch;
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>ch;
		a[i]=ch;
		def.insert(ch);
	}
	int k=def.size();//口袋妖怪的種類數
	int l=0,r=0,sum=0,ans=n;
	while(1)//兩層迴圈為了防止當r走到最右邊時，l還可以往左走的情況 
	{
		while(r<n&&sum<k)
		{
			if(num[a[r]]==0)//沒出現過 
				sum++;//以捕獲的口袋妖怪數量 
			num[a[r]]++; 
			r++; 
		}
		if(sum<k)
			break;
		ans=min(ans,r-l); 
		if(--num[a[l]]==0)
		{
			sum--;	
		}
			l++;
	}
	 cout<<ans<<endl;
	 return 0;
}