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 END {cout<<ans<<'\n'; return 0;}

int main(){
    cin.tie(nullptr); ios::sync_with_stdio(false);

    string s; cin>>s; string ans="-1"; int a[26]={},n=s.size();
    if(n==26){
        string s1=s;
        if(!next_permutation(s.begin(),s.end()))END
        ans="";
        for (int i = 0; i < n; i++){
            ans.push_back(s[i]);
            if(s[i]!=s1[i])END
        }
    }else{
    for (int i = 0; i < n; i++)a[s[i]-'a']++;
    for (int i = 0; i < 26; i++)if(!a[i]){
        s+=i+'a'; cout<<s<<'\n'; return 0;
    }
    }
}