Codeforces 888C:K-Dominant Character

You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.

You have to find minimum k such that there exists at least one k-dominant character.

Input

The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 100000).

Output

Print one number — the minimum value of k such that there exists at least one k-dominant character.

Examples

Input

abacaba

Output

2

Input

zzzzz

Output

1

Input

abcde

Output

3
读懂题意是第一步，每个字符存在一个子串长度k，每k长度的子串都包括这个字符，求最小k，计算每种字符的间距包括头和尾的最大值，然后求所有字符的最小值。
代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
char s[100001];
int last[26],ml[26];///last记录字母上一个位置，ml更新单个字母间距最大值
int main()
{
    scanf("%s",s);
    int i = -1;
    while(s[++ i])
    {
        if(s[last[s[i] - 'a']] == s[i])ml[s[i] - 'a'] = max(i - last[s[i] - 'a'],ml[s[i] - 'a']);///由于last初始0,需要判断上个位置是否是相同字符
        else ml[s[i] - 'a'] = i + 1;///位置从0开始 需要加1代表长度
        last[s[i] - 'a'] = i;
    }
    int ans = 200000;
    for(int i = 0;i < 26;i ++)
    {
        if(s[last[i]] == 'a' + i)///字符串中存在这个字符
        {
            if(ml[i])ml[i] = max(ml[i],(int)strlen(s) - last[i]);
            else ml[i] = max(last[i] + 1,(int)strlen(s) - last[i]);///如果ml为0，就计算字符到开头和结尾中的最大值
            ans = min(ans,ml[i]);
        }
    }
    printf("%d",ans);
}