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 Fst first
#define Snd second
#define RG register
#define mp make_pair
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned int UI;
typedef unsigned long long ULL;
template<typename T> inline void read(T& x) {
	char c = getchar();
	bool f = false;
	for (x = 0; !isdigit(c); c = getchar()) {
		if (c == '-') {
			f = true;
		}
	}
	for (; isdigit(c); c = getchar()) {
		x = x * 10 + c - '0';
	}
	if (f) {
		x = -x;
	}
}
template<typename T, typename... U> inline void read(T& x, U& ... y) {
	read(x), read(y...);
}
char s[110],g[110];
//#define rua
int main() {
//	ios::sync_with_stdio(false);
#ifdef rua
	freopen("GG.in","r",stdin);
#endif
  scanf("%s",s+1);
  memcpy(g,s,sizeof(s));
  int n=strlen(s+1),res=n;
  for(int i=0;i<26;++i) {
    int cnt=0;
    for(int j=1;j<=n;++j) if(s[j]-'a'==i) ++cnt;
    if(!cnt) continue;
    int m=n,t=0;
    while(1) {
      bool OK=true;
      for(int j=1;j<=m;++j) {
        if(s[j]-'a'!=i) {
          OK=false;
          break;
        }
      }
      if(OK) break;
      --m; ++t;
      for(int j=1;j<=m;++j) {
        if(s[j]-'a'!=i) {
          s[j]=s[j+1];
        }
      }
    }
    res=min(res,t);
    memcpy(s,g,sizeof(g));
  }
  printf("%d\n",res);
	return 0;
}