Description

Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The author of that text book, like other authors, is extremely fussy about the ideas, thus some ideas are covered more than once. Jessica think if she managed to read each idea at least once, she can pass the exam. She decides to read only one contiguous part of the book which contains all ideas covered by the entire book. And of course, the sub-book should be as thin as possible.

A very hard-working boy had manually indexed for her each page of Jessica's text-book with what idea each page is about and thus made a big progress for his courtship. Here you come in to save your skin: given the index, help Jessica decide which contiguous part she should read. For convenience, each idea has been coded with an ID, which is a non-negative integer.

Input

The first line of input is an integer P (1 ≤ P ≤ 1000000), which is the number of pages of Jessica's text-book. The second line contains P non-negative integers describing what idea each page is about. The first integer is what the first page is about, the second integer is what the second page is about, and so on. You may assume all integers that appear can fit well in the signed 32-bit integer type.

Output

Output one line: the number of pages of the shortest contiguous part of the book which contains all ideals covered in the book.

Sample Input

5
1 8 8 8 1

Sample Output

2

题意：求最短的 连续的子序列 使得出现所给序列中所有不同的数 输出长度
题解：记录左边界head=1
每次将右边位置向右移动一下，当出现a[i]==a[head]时尝试移动左边界
什么时候能移动呢？当最左边的数的计数大于1时移动 也就是最左边的数在当前区间出现次数>1
取满足条件的区间长度的最小值；

 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 //#include<bits/stdc++.h>
 8 #include<iostream>
 9 #include<cstring>
10 #include<cstdio>
11 #include<map>
12 #include<algorithm>
13 #include<cmath>
14 #define ll long long
15 #define PI acos(-1.0)
16 #define mod 1000000007
17 using namespace std;
18 int p;
19 int a[1000005];
20 int zong=0;
21 map<int,int>mp;
22 int main()
23 {
24     while(~scanf("%d",&p))
25     {
26         mp.clear();
27         zong=0;
28         for(int i=1; i<=p; i++)
29         {
30             scanf("%d",&a[i]);
31             if(mp[a[i]]==0)
32                 zong++;
33             mp[a[i]]++;
34         }
35         mp.clear();
36         int ans=0;
37         mp[a[1]]=1;
38         ans++;
39         int head=1;
40         int maxn=p;
41         for(int i=2; i<=p; i++)
42         {
43             if(mp[a[i]]==0)
44                 ans++;
45             mp[a[i]]++;
46             if(a[i]==a[head])
47             {
48                 while(mp[a[head]]>1)
49                 {
50                     mp[a[head]]--;
51                     head++;
52                 }
53             }
54             if(ans==zong)
55                 maxn=min(maxn,i-head+1);
56         }
57         printf("%d\n",maxn);
58     }
59     return 0;
60 }