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<bits/stdc++.h>
using namespace std;
const long long INF = 1e7+1;
const long long MOD = 1000000007;
const long double PI = 3.1415926;
#define FOR(i,r,n) for(ll i = (ll)(r); i < (ll)(n); i++)
#define RFOR(i,r,n) for(ll i=(ll)(n-1);i>=r;i--)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define ALL(x) x.begin(),x.end()
#define RALL(x) x.rbegin(),x.rend()
typedef long long int ll;
typedef vector<ll> vi;
typedef vector < pair<ll, ll > > vp;
typedef vector <string> vs;
typedef vector <char> vc;
typedef list <ll> lst;

ll n, k, ans = 0, sum = 0, cnt = 0;
string s;
char c;

/*--------------------template--------------------*/

int main()
{
	ans = 100;
	cin >> s;
	REP(i, s.size())
	{
		vi v;
		REP(j, s.size())
		{
			if (s[j] == s[i]) {
				v.push_back(j);
			}
		}
		sum = 0;
		sum = max(sum, v[0]);
		REP(j, v.size() - 1)
		{
			sum = max(sum, v[j + 1] - v[j] - 1);
		}
		sum = max<ll>(sum, (s.size() - 1 - v[v.size() - 1]));
		ans = min(ans, sum);
	}
	cout << ans << endl;
	return 0;
}