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<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < n; i++)
string s;
bool used[26];
int main(){
	cin >> s;
	if(s == "zyxwvutsrqponmlkjihgfedcba") return 0*puts("-1");
	else if(s.size() != 26){
		rep(i, s.size()) used[s[i]-'a'] = true;
		cout << s;
		rep(i,26) if(!used[i]) return 0*printf("%c\n", 'a'+i);
	}else{
		int t = 26, u;
		for(; s[t-1] > s[t]; t--) used[s[t]-'a'] = true;
		used[s[t--]-'a'] = true;
		for(u = s[t]-'a'; !used[u]; u++);
		rep(i,t) printf("%c", s[i]);
		printf("%c\n", 'a'+u);
	}
}