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;
string s;
bool used[26];
int main()
{
	cin>>s;
	for(int i=0;i<s.length();i++)
		used[s[i]-'a']=1;
	if(s.length()<26)
		for(int i=0;i<26;i++)
			if(!used[i])
			{
				cout<<s<<char(i+'a');
				return 0;
			}
	int id=-1;
	for(int i=s.length()-2;i>=0;i--)
		if(s[i]<s[i+1])
		{
			id=i;
			break;
		}
	if(id==-1)
	{
		puts("-1");
		return 0;
	}
	string t=s.substr(0,id);
	int idx2=id+1;
	for(int i=id+1;i<s.length();i++)
		if(s[i]<s[idx2]&&s[i]>s[id])
			idx2=i;
	t+=s[idx2];
	cout<<t;
	return 0;
}