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 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <string.h>

int min(int a,int b){
  if(a < b){
    return a;
  }
  return b;
}

int main(void){
  char s[100];
  scanf("%s", s);

  int N = strlen(s);
  int t[N];
  for (int i = 0; i < N; i++) {
    t[i] = s[i]-'a';
  }

  for (int i = 0; i < N-1; i++) {
    if(t[i] != t[i+1]) break;
    if(i == N-2){
      printf("0\n");
      return 0;
    }
  }

  int ans = N+1;
  int past[N], next[N];
  for (int i = 0; i < 26; i++) {
    int cnt = 1;
    for (int j = 0; j < N; j++) {
      past[j] = t[j];
    }
    while(1){
      for (int j = 0; j < N-cnt; j++) {
        if(past[j] == i || (j+1 < N && past[j+1] == i)){
          next[j] = i;
        }else {
          next[j] = past[j];
        }
      }

      for (int j = 0; j < N-cnt; j++) {
        if(next[j] != i) break;
        if(j == N-cnt-1) ans = min(ans, cnt);
      }

      for (int j = 0; j < N-cnt; j++) {
        past[j] = next[j];
      }
      cnt++;
      if(N == cnt) break;
    }
  }

  printf("%d\n", ans);

  return 0;
}