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>
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  for(int i=0;i<(n);++i)
#define REPr(i,n) for(int i=(n)-1;i>=0; --i)
#define FORq(i, m, n) for(int i = (m);i <= (n);++i)
#define FORqr(i, m , n) for(int i = (n);i >=(m);--i)
#define SCD(n) scanf("%d",&n)
#define SCD2(m,n) scanf("%d%d",&m,&n)
#define SCD3(m,n,k) scanf("%d%d%d",&m,&n,&k)
#define SCLLD(n) scanf("%lld",&n)
#define SCLLD2(m,n) scanf("%lld%lld",&m,&n)
#define SCLLD3(m,n,k) scanf("%lld%lld%lld",&m,&n,&k)
#define PB push_back
#define MP make_pair
#define ARSCD(A,N) REP(i,N){SCD(A[i]);}
#define ARSCD1(A,N) FORq(i,1,N){SCD(A[i]);}
#define VSCD(v,N) REP(i,N){int (x); SCD(x); v.PB(x);}
#define VSCLLD(v,N) REP(i,N){long long (x); SCLLD(x); v.PB(x);}
#define PRINTD(n) printf("%d\n",n)
#define PRINTLLD(n) printf("%lld\n",n)
#define DEBUG printf("%s\n","debug")
#define fst first
#define snd second
#define SIN(x,S) (S.count(x) != 0)
#define M0(x) memset(x,0,sizeof(x))
#define FILL(x,y) memset(x,y,sizeof(x))
#define MM(x) memset(x,-1,sizeof(x))
#define ALL(x) (x).begin(),(x).end()
using namespace std;
typedef pair<int,int> PII;
typedef pair<long long,long long> PLL;
typedef vector<int> VI;
typedef vector < VI > VVI;
typedef vector<long long> VL;
typedef long long ll;
typedef long long integer;
///////////////////////////////////////////////
const ll MOD = 1000000007;
ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
///////////////////////////////////////////////
int dx[2] = {0,1};
int dy[2] = {1,0};

int main(){
    string S;
    getline(cin,S);
    int ans = 1000;

    int N = S.length();



    REP(i,N){
        char target = S[i];
        string nowS = S;
        bool term = false;

        term = true;
        REP(k,nowS.length()){
            if (nowS[k] != target) term = false; 
        }

        if (term){
            PRINTD(0);
            return 0;
        }



        while(!term){
            string next = "";
            REP(j,nowS.length()-1){
                if (nowS[j] == target || nowS[j+1] == target){
                    next.push_back(target);
                }else{
                    next.push_back('*');
                }
            }

            term = true;
            REP(k,next.length()){
               if (next[k] != target) term = false; 
            }

            nowS = next;
        }

        if (ans >= N - nowS.length()) ans = N - nowS.length();
        //printf("target = %c, %s \n",target,)

    }

    cout << ans << endl;

}