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;

#define int long long
#define fi first
#define se second
#define otp(n) cout << n;
#define ct(n) cout << n << endl;
#define cts(n) cout << n << " ";
#define cte cout << endl;
#define ctv(v) rep(i,(v).size()) ct(v[i])
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define repo(i,o,n) for(int i=o;i<n;i++)
#define repm(i,n) for(int i=n-1;i>=0;i--)
#define repv(i,v) for(int i=0;i<(v).size();i++)
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define sperase(v,n) (v).erase(remove(all(v), n), (v).end());
#define vdelete(v) (v).erase(unique(all(v)), (v).end());
#define pb(n) push_back(n);

int ans=100;
string s;

main() {
    cin >> s;
    rep(i,26) {
        int cnt=0,cntmax=0;
        rep(j,s.size()) {
            if ('a'+i == s[j]) cntmax = max(cntmax,cnt),cnt=0;
            else cnt++;
        } cntmax = max(cntmax,cnt);
        ans = min(ans,cntmax);
    }
    ct(ans)
}