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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define rep(i,n) for (int i=0;i<(int)(n);i++)
#define codefor int test;scanf("%d",&test);while(test--)
#define INT(...) int __VA_ARGS__;in(__VA_ARGS__)
#define LL(...) ll __VA_ARGS__;in(__VA_ARGS__)
#define yn(ans) if(ans)printf("Yes\n");else printf("No\n")
#define YN(ans) if(ans)printf("YES\n");else printf("NO\n")
#define vector1d(type,name,...) vector<type>name(__VA_ARGS__)
#define vector2d(type,name,h,...) vector<vector<type>>name(h,vector<type>(__VA_ARGS__))
#define umap unordered_map
#define uset unordered_set
using namespace std;
using ll = long long;
const int MOD=1000000007;
//入力系
void scan(int& a){scanf("%d",&a);}
void scan(long long& a){scanf("%lld",&a);}
template<class T> void scan(T& a){cin>>a;}
template<class T> void scan(vector<T>& vec){for(auto&& it:vec)scan(it);}
void in(){}
template <class Head, class... Tail> void in(Head& head, Tail&... tail){scan(head);in(tail...);}
//出力系
void print(const int& a){printf("%d",a);}
void print(const long long& a){printf("%lld",a);}
void print(const double& a){printf("%.15lf",a);}
template<class T> void print(const T& a){cout<<a;}
template<class T> void print(const vector<T>& vec){if(vec.empty())return;print(vec[0]);for(auto it=vec.begin();++it!= vec.end();){putchar(' ');print(*it);}}
void out(){putchar('\n');}
template<class T> void out(const T& t){print(t);putchar('\n');}
template <class Head, class... Tail> void out(const Head& head,const Tail&... tail){print(head);putchar(' ');out(tail...);}
//デバッグ系
template<class T> void dprint(const T& a){cerr<<a;}
template<class T> void dprint(const vector<T>& vec){if(vec.empty())return;cerr<<vec[0];for(auto it=vec.begin();++it!= vec.end();){cerr<<" "<<*it;}}
void debug(){cerr<<endl;}
template<class T> void debug(const T& t){dprint(t);cerr<<endl;}
template <class Head, class... Tail> void debug(const Head& head, const Tail&... tail){dprint(head);cerr<<" ";debug(tail...);}
ll intpow(ll a, ll b){ ll ans = 1; while(b){ if(b & 1) ans *= a; a *= a; b /= 2; } return ans; }
ll modpow(ll a, ll b, ll p){ ll ans = 1; while(b){ if(b & 1) (ans *= a) %= p; (a *= a) %= p; b /= 2; } return ans; }
ll updivide(ll a,ll b){if(a%b==0) return a/b;else return (a/b)+1;}

int main(){
    string s,t;
    cin>>s;
    char c='a';
    int ans=1000,d=0;
    set<char> kind;
    rep(i,s.length())kind.insert(s[i]);
    for(auto key:kind){
        t=s;
        d=0;
        while(count(all(t),key)!=t.length()){
            d++;
            for(int i=t.length()-1;i>=0;i--){
                if(t[i]!=key){
                    t.erase(t.begin()+i);
                    for(int j=0;j<i;j++){
                        if(j+1<i&&t[j+1]==key){
                            t[j]=key;
                        }
                    }
                    goto skip;
                }
            }
            skip: ;
            //debug(t);
        }
        ans=min(ans,d);
    }
    out(ans);
}