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 OUT(x) cout << #x << " = " << x << endl
#define rep(i, n)             for (int (i) = 0; (i) < (int)(n); (i)++)
#define rer(i, l, r)          for (int (i) = (int)(l); (i) <= (int)(r); (i)++)
#define reu(i, l, r)          for (int (i) = (int)(l); (i) < (int)(r); (i)++)
#define all(x)                (x).begin(), (x).end()
#define rall(x)               (x).rbegin(), (x).rend()
template<typename T> void pv(T a, T b) { for (T i = a; i != b; i ++) cout << *i << " "; cout << endl; }
template<typename T, typename U> inline void amin(T &x, U y) { if (y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if (x < y) x = y; }
int in() { int x; scanf("%d", &x); return x; }
long long lin() { long long x; scanf("%lld", &x); return x; }

static const int INF = 0x3f3f3f3f;

signed main() { 
        string s;
        cin >> s;
        int ans = INF;
        rep(i, 26) {
                int res = 0;
                int cnt = 0;
                bool f = false;
                rep(j, s.size()) {
                        cnt ++;
                        if (s[j] == 'a' + i) {
                                f = true;
                                amax(res, cnt - 1);
                                cnt = 0;
                        }
                }
                if (f) amax(res, cnt);
                if (f) amin(ans, res);
        }
        cout << ans << endl;
        return 0;
}