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 sc = new Scanner(System.in);
		String s = sc.next();
		sc.close();
		int l = s.length();

		if(l == 26) {
			int[] p = new int[l];
			for(int i = 0; i < l; i++) {
				p[i] = s.charAt(i) - 'a';
			}
			if(next_permutation(p)) {
				for(int i = 0; i < 26; i++) {
					if(s.charAt(i) - 'a' == p[i]) {
						System.out.print(s.charAt(i));
					}else {
						System.out.println((char)(p[i] + 'a'));
						break;
					}
				}
			}else {
				System.out.println(-1);
			}
		}else {
			int[] a = new int[26];
			for(int i = 0; i < l; i++) {
				a[s.charAt(i) - 'a']++;
			}
			for(int i = 0; i < 26; i++) {
				if(a[i] == 0) {
					System.out.print(s);
					System.out.println((char)(i + 'a'));
					break;
				}
			}
		}
	}
	static boolean next_permutation(int[] p) {
		for(int a = p.length - 2; a >= 0; --a) {
			if(p[a] < p[a + 1]) {
				for(int b = p.length - 1;; --b) {
					if(p[b] > p[a]) {
						int t = p[a];
						p[a] = p[b];
						p[b] = t;
						for(++a, b = p.length - 1; a < b; ++a, --b) {
							t = p[a];
							p[a] = p[b];
							p[b] = t;
						}
						return true;
					}
				}
			}
		}
		return false;
	}
}