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<iostream>
using namespace std;
string S;
bool a[26];
int main(){
    cin>>S;
    if(S=="zyxwvutsrqponmlkjihgfedcba"){cout<<-1<<endl;return 0;}
    for(int i=0;i<S.size();i++)a[S[i]-'a']=true;

    if(S.size()!=26)for(int i=0;i<26;i++)if(a[i]==false){cout<<S+char('a'+i)<<endl;return 0;}
    for(int i=0;i<S.size();i++){
        for(int c=S[S.size()-1-i]-'a';c<26;c++){
            if(a[c]==false){
                S[S.size()-1-i]=char('a'+c);
                cout<<S.substr(0,S.size()-i)<<endl;
                return 0;
            }
        }
        a[S[S.size()-1-i]-'a']=false;
    }
    return 0;
}