Problem:
Score : 300 points

Problem Statement

Snuke can change a string t of length N into a string t' of length N - 1 under the following rule:

For each i (1 ≤ i ≤ N - 1), the i-th character of t' must be either the i-th or (i + 1)-th character of t.

There is a string s consisting of lowercase English letters.
Snuke's objective is to apply the above operation to s repeatedly so that all the characters in s are the same.
Find the minimum necessary number of operations.

Constraints

1 ≤ |s| ≤ 100

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

Print the minimum necessary number of operations to achieve the objective.

Sample Input 1

serval

Sample Output 1

3

One solution is: serval → srvvl → svvv → vvv.

Sample Input 2

jackal

Sample Output 2

2

One solution is: jackal → aacaa → aaaa.

Sample Input 3

zzz

Sample Output 3

0

All the characters in s are the same from the beginning.

Sample Input 4

whbrjpjyhsrywlqjxdbrbaomnw

Sample Output 4

8

In 8 operations, he can change s to rrrrrrrrrrrrrrrrrr.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<utility>
#include<numeric>
#include<vector>
#include<map>
#include<unordered_map>
#include<set>
#include<tuple>
#include<stack>
#include<queue>
#include<tuple>
#include<functional>
#include<iterator>
#include<cmath>
#include<cctype>

using namespace std;

typedef long long ll;
typedef pair<int,int> P;

const int INF = 1e9;
const ll LINF = 1e18;
const int MOD = INF+7;

struct edge{int to,cost;};

bool cs[26];

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);

	string tmp;
	cin >> tmp;
	for(int i=0;i<tmp.size();i++){
		cs[tmp[i]-'a']=true;
	}

	string s;
	int ans=INF;
	for(char c='a';c<='z';c++){
		if(!cs[c-'a']) continue;
		bool yet=true;
		int cnt=0;
		s = tmp;
		int len=s.size();

		while(yet){
			string ss(100,'A');
			yet=false;
			for(int i=0;i<len-1;i++){
				if(s[i] == c || s[i+1] == c) ss[i]=c;
				else{
					ss[i] = s[i];
					yet=true;
				}
			}
			if(cnt==0 && !yet){
				cout << "0\n";
				return 0;
			}
			cnt++;
			s = ss;
			len--;
		}
		ans = min(ans,cnt);
	}
	cout << ans << "\n";

	return 0;
}