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 <algorithm>
#include <array>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <vector>

#define REP(X, Y, Z) for (ll(X) = (Y); (X) < (Z); (X)++)
#define REV(X, Y, Z) for (ll(X) = (Y); (X) > (Z); (X)--)

using namespace std;
using ll = long long;
using dll = deque<ll>;
using pll = pair<ll, ll>;
using qll = queue<ll>;
using usll = unordered_set<ll>;
using umll = unordered_map<ll, ll>;
using sll = stack<int>;
using vll = vector<ll>;
using vvll = vector<vll>;

constexpr ll kMax = 1e12 + 1;
constexpr ll kMod = 1e9 + 7;
constexpr long double kPI = 3.14159265358979323846;

ll gcd(ll a, ll b) {
  if (a < b) {
    return gcd(b, a);
  }
  ll c = a % b;
  while (c != 0) {
    a = b;
    b = c;
    c = a % b;
  }
  return b;
}

ll lcm(ll a, ll b) {
  return a / gcd(a, b) * b;
}

void init() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}

int main() {
  init();

  string S;
  cin >> S;

  ll ans = kMax;
  for (char c = 'a'; c <= 'z'; c++) {
    string st = S;
    REV(j, S.size(), -1) {
      ll ss = 0;
      REP(k, 0, j) {
        if (st[k] == c) {
          ss++;
        }
        if (k + 1 < S.size() && st[k + 1] == c) {
          st[k] = c;
        }
      }
      if (ss == j) {
        ans = min(ans, (ll) S.size() - j);
        break;
      }
    }
  }
  cout << ans << endl;

  return 0;
}