Problem:
Score : 300 points

Problem Statement

Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.

A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, atcoder, zscoder and agc are diverse words while gotou and connect aren't diverse words.

Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.

Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.

Constraints

1 \leq |S| \leq 26

S is a diverse word.

Input

Input is given from Standard Input in the following format:

S

Output

Print the next word that appears after S in the dictionary, or -1 if it doesn't exist.

Sample Input 1

atcoder

Sample Output 1

atcoderb

atcoderb is the lexicographically smallest diverse word that is lexicographically larger than atcoder. Note that atcoderb is lexicographically smaller than b.

Sample Input 2

abc

Sample Output 2

abcd

Sample Input 3

zyxwvutsrqponmlkjihgfedcba

Sample Output 3

-1

This is the lexicographically largest diverse word, so the answer is -1.

Sample Input 4

abcdefghijklmnopqrstuvwzyx

Sample Output 4

abcdefghijklmnopqrstuvx

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

s = gets.chomp.split("")
alphabets = ("a".."z").to_a
if addition = alphabets.find { |a| !s.include?(a) }
  s << addition
  puts s.join
else
  previous_order = s.first.ord
  must_change = nil
  s[1..-1].each.with_index(1) do |c, index|
    if previous_order + 1 < c.ord
      must_change = index - 1
    end
    previous_order = c.ord
  end
  if must_change
    s[0...must_change].each do |c|
      alphabets.delete(c)
    end
    new_s = s[0...must_change]
    new_s << alphabets[1]
    puts new_s.join
  else
    last_order = s.last.ord
    must_change_2 = nil
    s[0...-1].reverse.each_with_index do |c, index|
      if last_order > c.ord
        must_change_2 = index
        break
      else
        last_order = c.ord
      end
    end
    if must_change_2
      must_change_2 = s.size - must_change_2 - 2
      s[must_change_2], s[must_change_2 + 1] = s[must_change_2 + 1], s[must_change_2]
      puts s[0..must_change_2].join
    else
      puts -1
    end
  end
end