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.io.*;
import java.math.*;
import java.util.*;
public class Main { 

	public static void main(String[] args) { //GIMME AC
		Scanner input = new Scanner(System.in); 
		String S = input.next();
		int N = S.length();
		String ans = "";
		boolean[] v = new boolean[200];
		if (N<26) {
			ans+=S;
			for (int i = 0; i < N; i++) {
				v[(int)(S.charAt(i))]=true;
			}
			for (int i = (int)('a'); i <= (int)('z'); i++) {
				if (!v[i]) {
					ans+=Character.toString((char)(i));
					break;
				} 
			}
		}else {
			int L = -1; //L=0 is still valid dawg
			for (int R = N-2; R >= 0; R--) {
				if ((S.charAt(R))>(S.charAt(R+1))) continue;
				else {
					L=R;
					break;
				}
			}
			if (L==-1) ans="-1";
			else {
				ans+=S.substring(0,L);
				char cmp = S.charAt(L);
				for (int i = 0; i < L; i++) {
					v[(int)(ans.charAt(i))]=true;
				}
				for (int i = (int)('a'); i <= (int)('z'); i++) {
					if (!v[i]&&i>(int)(cmp)) {
						ans+=Character.toString((char)(i));
						break;
					}
				}
			}
		}
		System.out.println(ans);
	}
}