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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
// #include <boost/multiprecision/cpp_int.hpp> 

#define gc getchar//_unlocked
#define pc putchar//_unlocked
#define ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define pp pair<int,int>
#define ppl pair<ll,ll>
#define bigint boost::multiprecision::cpp_int
#define finp ios_base::sync_with_stdio(0);cin.tie(0);
#define bc __builtin_popcountll
#define afor(i,a,b) for(int i=a;i<=b;++i)
#define bfor(i,a,b) for(int i=a;i>=b;--i)
#define vi vector<int>
#define vpp vector<pp>
#define vll vector<ll>

using namespace std;
using namespace __gnu_pbds;

char putnb[20];
void putn(ll n) {if(!n)pc('0');if(n<0)pc('-'),n=0-n;int pi=0;while(n)putnb[pi++]=(n%10)+'0',n/=10;while(pi)pc(putnb[--pi]);}
void sci(int *x) {register char c = gc();*x = 0;for(; (c<48)||(c>57);c = gc());for(; (c>47)&&(c<58);c = gc())*x = (int)((((*x)<<1) + ((*x)<<3)) + c - 48);}
void scll(ll *x)  {register char c = gc();*x = 0;for(; (c<48)||(c>57);c = gc());for(; (c>47)&&(c<58);c = gc())*x =  (ll)((((*x)<<1) + ((*x)<<3)) + c - 48);}
ll fp(ll a,ll b,ll c) {if(b==0)return 1%c; if(b==1)return a%c; ll ret=fp(a,b/2,c); ret=(ret*ret)%c; if(b&1)ret=(ret*a)%c; return ret;}

const ll mod=1e9 +7;
const ll mod2=1999999973;
const ll inf=1e18;
const int infs=1e9 + 1000;
const int N=100000;
const long double PI = acos(-1);

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;


int main()
{
  finp;
  string s;
  cin>>s;
  int n = s.length();
  int ans = infs;
  afor(j,0,25)
  {
    string ss = s;
    int curr = 0;
    int sz = n;

    while(1)
    {
      bool f= true;
      afor(i,0,sz-1)
      {
        if(ss[i] != (j+'a'))f = false;
      }
      if(f)break;
      curr++;
      afor(i,0,sz-2)
      {
        if(ss[i]==(j+'a') || ss[i+1]==(j+'a'))ss[i] = j + 'a';
      }
      ss.pop_back();
      sz--;
    }
    ans = min(ans,curr);
  }
  assert(ans != infs);
  cout<<ans;
return 0;
}