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
import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        char[] alps = "abcdefghijklmnopqrstuvwxyz".toCharArray();
        char[] s = sc.next().toCharArray();
        for(char c : alps){
            boolean bool = false;
            for(int i = 0; i < s.length; i++){
                if(s[i] == c){
                    bool = true;
                    break;
                }
            }
            if(!bool){
                System.out.println(new String(s) + c);
                return;
            }
        }
        ArrayList<Character> cl = new ArrayList<Character>();
        for(int i = 25; i > 0; i--){
            cl.add(s[i]);
            if(s[i] - s[i-1] > 0){
                Collections.sort(cl);
                for(char c : cl){
                    if(s[i-1] - c < 0){
                        s[i-1] = c;
                        String ans = new String(Arrays.copyOfRange(s,0,i));
                        System.out.println(ans);
                        return;
                    } 
                }
            }
        }
        System.out.println(-1);
    }
}