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.

Examples

Input

atcoder


Output

atcoderb


Input

abc


Output

abcd


Input

zyxwvutsrqponmlkjihgfedcba


Output

-1


Input

abcdefghijklmnopqrstuvwzyx


Output

abcdefghijklmnopqrstuvx
#include <bits/stdc++.h>
using namespace std;
 
int main() {
  string s;
  cin>>s;
  int n=s.size();
  int a[26]={};
  
  if(s=="zyxwvutsrqponmlkjihgfedcba"){
    cout<<-1;
    return 0;
  }
  
  else if(n!=26){
    for(int i=0;i<n;i++)
      a[s[i]-'a']++;
    for(int i=0;i<26;i++)
      if(a[i]==0){
        s+=char(i+'a');
        break;
      }
    cout<<s;
    return 0;
  }
  
  
  else if(n==26){
    char ss[26];
    for(int j=0;j<26;j++)
      ss[j]=s[j];
    next_permutation(ss,ss+n);
    for(int i=0;i<n;i++){
      cout<<char(ss[i]);
      if(s[i]!=ss[i])
        break;
    }
  }

  
  
}