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<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<queue>
#include<vector>
using namespace std;
int main()
{
  string s;
  cin>>s;
  if(s=="zyxwvutsrqponmlkjihgfedcba"){
    cout<<"-1";
    return 0;
  }
  s+='a'-1;
  for(int i=s.size()-1;i>=0;i--)
    while(s[i]<'z'){
      s[i]++;
      bool f=true;
      for(int j=0;j<i;j++)
        if(s[i]==s[j]){
          f=false;
          break;
        }
      if(f){
        for(int j=0;j<=i;j++)
          cout<<s[j];
        return 0;
      }
    }
}