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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>

using lli = long long int;

#define REP(i, n) for(int (i) = 0; (i) < (n); (i)++)

template <class S, class T> inline void chmax(S &a, T b){if(a < b) a = b;}
template <class S, class T> inline void chmin(S &a, T b){if(a > b) a = b;}

constexpr int MAX_N = 100;
constexpr int ALPHABET = 'z' - 'a' + 1;
constexpr int INF = 1e9 + 1;
char str[MAX_N+1];
std::vector<int> A[ALPHABET];

int main(void){
    scanf("%s", str); int n = 0;
    while(str[n] != '\0'){
        A[str[n] - 'a'].push_back(n);
        n++;
    }
    int res = INF;
    REP(i, ALPHABET) if(A[i].size() != 0){
        const int s = A[i].size();
        int mx = -1;
        chmax(mx, A[i][0]);
        REP(j, s-1){
            int diff = A[i][j+1] - A[i][j];
            chmax(mx, diff - 1);
            if(i + 'a' == 'r') printf("Diff => %d\n", diff);
        }
        chmax(mx, n - A[i][s-1] - 1);
        chmin(res, mx);
    }
    printf("%d\n", res);
    return 0;
}