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; 

typedef long long ll; 

int main(){
    string s; cin>>s; 
    vector<int> vv(26,0); 
    for(int i=0;i<s.size();i++) { vv[s[i]-97]++; }
    if(s.size()<26){
        cout<<s; 
        for(int i=0;i<26;i++) { if(vv[i]==0) { cout<<(char)(i+97); return 0;} }
    }
    for(int i=25;i>=1;i--){
        if(s[i] > s[i-1]) { cout<<s.substr(0,i-1); char c=s[i]; for(int j=i+1;j<s.size();j++) { if(s[j]<c && s[j]>s[i-1]) c=s[j]; } cout<<c;  return 0; }
    }
    cout<<-1; 

	return 0; 

}

