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.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        String S = scan.next();
        System.out.println(solve(S));
    }
    private static String solve(String S) {
        char[] list = S.toCharArray();
        int[] use_chars = new int[26];
        for (int i = list.length - 1; 0 <= i; i--) {
            use_chars[(int)list[i]-97] += 1;
        }
        if(list.length < 26) {
            for(int i = 0; i < 26; i++) {
                if(use_chars[i] == 0) {
                    return S + (char) (i+97);
                }
            }
        }
        for (int i = list.length - 1; 0 <= i; i--) {
            int current = list[i] - 97;
            for(int j = current + 1; j < 26; j++) {
                if(use_chars[j] == 0) {
                    return S.substring(0, i) + (char) (j+97);
                }
            }
            use_chars[current] -= 1;
        }
        return "-1";

    }

}
